Fix LaTeX in math_differential_calculus.ipynb
Joaquín Ruales
GitHub
parent
43f8795bcc
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a8f393cf34
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@ -439,16 +439,16 @@
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"Let's look at a concrete example. Let's see if we can determine what the slope of the $y=x^2$ curve is, at any point $\\mathrm{A}$ (try to understand each line, I promise it's not that hard):\n",
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"\n",
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"$\n",
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"\\begin{split}\n",
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"f'(x_\\mathrm{A}) \\, && = \\underset{x_\\mathrm{B} \\to x_\\mathrm{A}}\\lim\\dfrac{f(x_\\mathrm{B}) - f(x_\\mathrm{A})}{x_\\mathrm{B} - x_\\mathrm{A}} \\\\\n",
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"&& = \\underset{x_\\mathrm{B} \\to x_\\mathrm{A}}\\lim\\dfrac{{x_\\mathrm{B}}^2 - {x_\\mathrm{A}}^2}{x_\\mathrm{B} - x_\\mathrm{A}} \\quad && \\text{since } f(x) = x^2\\\\\n",
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"&& = \\underset{x_\\mathrm{B} \\to x_\\mathrm{A}}\\lim\\dfrac{(x_\\mathrm{B} - x_\\mathrm{A})(x_\\mathrm{B} + x_\\mathrm{A})}{x_\\mathrm{B} - x_\\mathrm{A}}\\quad && \\text{since } {x_\\mathrm{A}}^2 - {x_\\mathrm{B}}^2 = (x_\\mathrm{A}-x_\\mathrm{B})(x_\\mathrm{A}+x_\\mathrm{B})\\\\\n",
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"&& = \\underset{x_\\mathrm{B} \\to x_\\mathrm{A}}\\lim(x_\\mathrm{B} + x_\\mathrm{A})\\quad && \\text{since the two } (x_\\mathrm{B} - x_\\mathrm{A}) \\text{ cancel out}\\\\\n",
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"&& = \\underset{x_\\mathrm{B} \\to x_\\mathrm{A}}\\lim x_\\mathrm{B} \\, + \\underset{x_\\mathrm{B} \\to x_\\mathrm{A}}\\lim x_\\mathrm{A}\\quad && \\text{since the limit of a sum is the sum of the limits}\\\\\n",
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"&& = x_\\mathrm{A} \\, + \\underset{x_\\mathrm{B} \\to x_\\mathrm{A}}\\lim x_\\mathrm{A} \\quad && \\text{since } x_\\mathrm{B}\\text{ approaches } x_\\mathrm{A} \\\\\n",
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"&& = x_\\mathrm{A} + x_\\mathrm{A} \\quad && \\text{since } x_\\mathrm{A} \\text{ remains constant when } x_\\mathrm{B}\\text{ approaches } x_\\mathrm{A} \\\\\n",
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"&& = 2 x_\\mathrm{A}\n",
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"\\end{split}\n",
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"\\begin{align*}\n",
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"f'(x_\\mathrm{A}) \\, & = \\underset{x_\\mathrm{B} \\to x_\\mathrm{A}}\\lim\\dfrac{f(x_\\mathrm{B}) - f(x_\\mathrm{A})}{x_\\mathrm{B} - x_\\mathrm{A}} \\\\\n",
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"& = \\underset{x_\\mathrm{B} \\to x_\\mathrm{A}}\\lim\\dfrac{{x_\\mathrm{B}}^2 - {x_\\mathrm{A}}^2}{x_\\mathrm{B} - x_\\mathrm{A}} \\quad & \\text{since } f(x) = x^2\\\\\n",
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"& = \\underset{x_\\mathrm{B} \\to x_\\mathrm{A}}\\lim\\dfrac{(x_\\mathrm{B} - x_\\mathrm{A})(x_\\mathrm{B} + x_\\mathrm{A})}{x_\\mathrm{B} - x_\\mathrm{A}}\\quad & \\text{since } {x_\\mathrm{A}}^2 - {x_\\mathrm{B}}^2 = (x_\\mathrm{A}-x_\\mathrm{B})(x_\\mathrm{A}+x_\\mathrm{B})\\\\\n",
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"& = \\underset{x_\\mathrm{B} \\to x_\\mathrm{A}}\\lim(x_\\mathrm{B} + x_\\mathrm{A})\\quad & \\text{since the two } (x_\\mathrm{B} - x_\\mathrm{A}) \\text{ cancel out}\\\\\n",
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"& = \\underset{x_\\mathrm{B} \\to x_\\mathrm{A}}\\lim x_\\mathrm{B} \\, + \\underset{x_\\mathrm{B} \\to x_\\mathrm{A}}\\lim x_\\mathrm{A}\\quad & \\text{since the limit of a sum is the sum of the limits}\\\\\n",
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"& = x_\\mathrm{A} \\, + \\underset{x_\\mathrm{B} \\to x_\\mathrm{A}}\\lim x_\\mathrm{A} \\quad & \\text{since } x_\\mathrm{B}\\text{ approaches } x_\\mathrm{A} \\\\\n",
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"& = x_\\mathrm{A} + x_\\mathrm{A} \\quad & \\text{since } x_\\mathrm{A} \\text{ remains constant when } x_\\mathrm{B}\\text{ approaches } x_\\mathrm{A} \\\\\n",
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"& = 2 x_\\mathrm{A}\n",
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"\\end{align*}\n",
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"$\n",
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"\n",
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"That's it! We just proved that the slope of $y = x^2$ at any point $\\mathrm{A}$ is $f'(x_\\mathrm{A}) = 2x_\\mathrm{A}$. What we have done is called **differentiation**: finding the derivative of a function."
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@ -517,14 +517,14 @@
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"Okay! Now let's use this new definition to find the derivative of $f(x) = x^2$ at any point $x$, and (hopefully) we should find the same result as above (except using $x$ instead of $x_\\mathrm{A}$):\n",
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"\n",
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"$\n",
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"\\begin{split}\n",
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"f'(x) \\, && = \\underset{\\epsilon \\to 0}\\lim\\dfrac{f(x + \\epsilon) - f(x)}{\\epsilon} \\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{(x + \\epsilon)^2 - {x}^2}{\\epsilon} \\quad && \\text{since } f(x) = x^2\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{{x}^2 + 2x\\epsilon + \\epsilon^2 - {x}^2}{\\epsilon}\\quad && \\text{since } (x + \\epsilon)^2 = {x}^2 + 2x\\epsilon + \\epsilon^2\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{2x\\epsilon + \\epsilon^2}{\\epsilon}\\quad && \\text{since the two } {x}^2 \\text{ cancel out}\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim \\, (2x + \\epsilon)\\quad && \\text{since } 2x\\epsilon \\text{ and } \\epsilon^2 \\text{ can both be divided by } \\epsilon\\\\\n",
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"&& = 2 x\n",
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"\\end{split}\n",
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"\\begin{align*}\n",
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"f'(x) \\, & = \\underset{\\epsilon \\to 0}\\lim\\dfrac{f(x + \\epsilon) - f(x)}{\\epsilon} \\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{(x + \\epsilon)^2 - {x}^2}{\\epsilon} \\quad & \\text{since } f(x) = x^2\\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{{x}^2 + 2x\\epsilon + \\epsilon^2 - {x}^2}{\\epsilon}\\quad & \\text{since } (x + \\epsilon)^2 = {x}^2 + 2x\\epsilon + \\epsilon^2\\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{2x\\epsilon + \\epsilon^2}{\\epsilon}\\quad & \\text{since the two } {x}^2 \\text{ cancel out}\\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim \\, (2x + \\epsilon)\\quad & \\text{since } 2x\\epsilon \\text{ and } \\epsilon^2 \\text{ can both be divided by } \\epsilon\\\\\n",
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"& = 2 x\n",
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"\\end{align*}\n",
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"$\n",
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"\n",
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"Yep! It works out."
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@ -705,13 +705,13 @@
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"One very important rule is that **the derivative of a sum is the sum of the derivatives**. More precisely, if we define $f(x) = g(x) + h(x)$, then $f'(x) = g'(x) + h'(x)$. This is quite easy to prove:\n",
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"\n",
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"$\n",
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"\\begin{split}\n",
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"f'(x) && = \\underset{\\epsilon \\to 0}\\lim\\dfrac{f(x+\\epsilon) - f(x)}{\\epsilon} && \\quad\\text{by definition}\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{g(x+\\epsilon) + h(x+\\epsilon) - g(x) - h(x)}{\\epsilon} && \\quad \\text{using }f(x) = g(x) + h(x) \\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{g(x+\\epsilon) - g(x) + h(x+\\epsilon) - h(x)}{\\epsilon} && \\quad \\text{just moving terms around}\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{g(x+\\epsilon) - g(x)}{\\epsilon} + \\underset{\\epsilon \\to 0}\\lim\\dfrac{h(x+\\epsilon) - h(x)}{\\epsilon} && \\quad \\text{since the limit of a sum is the sum of the limits}\\\\\n",
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"&& = g'(x) + h'(x) && \\quad \\text{using the definitions of }g'(x) \\text{ and } h'(x)\n",
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"\\end{split}\n",
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"\\begin{align*}\n",
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"f'(x) & = \\underset{\\epsilon \\to 0}\\lim\\dfrac{f(x+\\epsilon) - f(x)}{\\epsilon} & \\quad\\text{by definition}\\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{g(x+\\epsilon) + h(x+\\epsilon) - g(x) - h(x)}{\\epsilon} & \\quad \\text{using }f(x) = g(x) + h(x) \\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{g(x+\\epsilon) - g(x) + h(x+\\epsilon) - h(x)}{\\epsilon} & \\quad \\text{just moving terms around}\\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{g(x+\\epsilon) - g(x)}{\\epsilon} + \\underset{\\epsilon \\to 0}\\lim\\dfrac{h(x+\\epsilon) - h(x)}{\\epsilon} & \\quad \\text{since the limit of a sum is the sum of the limits}\\\\\n",
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"& = g'(x) + h'(x) & \\quad \\text{using the definitions of }g'(x) \\text{ and } h'(x)\n",
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"\\end{align*}\n",
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"$"
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]
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},
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@ -1213,18 +1213,18 @@
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"$\n",
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"\\mathbf{J}_\\mathbf{f}(\\mathbf{x}_\\mathbf{A}) = \\begin{pmatrix}\n",
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"\\dfrac{\\partial f_1}{\\partial x_1}(\\mathbf{x}_\\mathbf{A})\n",
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"&& \\dfrac{\\partial f_1}{\\partial x_2}(\\mathbf{x}_\\mathbf{A})\n",
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"&& \\dots\n",
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"&& \\dfrac{\\partial f_1}{\\partial x_n}(\\mathbf{x}_\\mathbf{A})\\\\\n",
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"& \\dfrac{\\partial f_1}{\\partial x_2}(\\mathbf{x}_\\mathbf{A})\n",
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"& \\dots\n",
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"& \\dfrac{\\partial f_1}{\\partial x_n}(\\mathbf{x}_\\mathbf{A})\\\\\n",
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"\\dfrac{\\partial f_2}{\\partial x_1}(\\mathbf{x}_\\mathbf{A})\n",
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"&& \\dfrac{\\partial f_2}{\\partial x_2}(\\mathbf{x}_\\mathbf{A})\n",
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"&& \\dots\n",
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"&& \\dfrac{\\partial f_2}{\\partial x_n}(\\mathbf{x}_\\mathbf{A})\\\\\n",
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"\\vdots && \\vdots && \\ddots && \\vdots \\\\\n",
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"& \\dfrac{\\partial f_2}{\\partial x_2}(\\mathbf{x}_\\mathbf{A})\n",
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"& \\dots\n",
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"& \\dfrac{\\partial f_2}{\\partial x_n}(\\mathbf{x}_\\mathbf{A})\\\\\n",
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"\\vdots & \\vdots & \\ddots & \\vdots \\\\\n",
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"\\dfrac{\\partial f_m}{\\partial x_1}(\\mathbf{x}_\\mathbf{A})\n",
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"&& \\dfrac{\\partial f_m}{\\partial x_2}(\\mathbf{x}_\\mathbf{A})\n",
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"&& \\dots\n",
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"&& \\dfrac{\\partial f_m}{\\partial x_n}(\\mathbf{x}_\\mathbf{A})\n",
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"& \\dfrac{\\partial f_m}{\\partial x_2}(\\mathbf{x}_\\mathbf{A})\n",
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"& \\dots\n",
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"& \\dfrac{\\partial f_m}{\\partial x_n}(\\mathbf{x}_\\mathbf{A})\n",
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"\\end{pmatrix}\n",
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"$\n",
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"\n",
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@ -1257,18 +1257,18 @@
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"$\n",
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"\\mathbf{H}_f(\\mathbf{x}_\\mathbf{A}) = \\begin{pmatrix}\n",
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"\\dfrac{\\partial^2 f}{\\partial {x_1}^2}(\\mathbf{x}_\\mathbf{A})\n",
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"&& \\dfrac{\\partial^2 f}{\\partial x_1\\, \\partial x_2}(\\mathbf{x}_\\mathbf{A})\n",
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"&& \\dots\n",
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"&& \\dfrac{\\partial^2 f}{\\partial x_1\\, \\partial x_n}(\\mathbf{x}_\\mathbf{A})\\\\\n",
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"& \\dfrac{\\partial^2 f}{\\partial x_1\\, \\partial x_2}(\\mathbf{x}_\\mathbf{A})\n",
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"& \\dots\n",
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"& \\dfrac{\\partial^2 f}{\\partial x_1\\, \\partial x_n}(\\mathbf{x}_\\mathbf{A})\\\\\n",
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"\\dfrac{\\partial^2 f}{\\partial x_2\\,\\partial x_1}(\\mathbf{x}_\\mathbf{A})\n",
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"&& \\dfrac{\\partial^2 f}{\\partial {x_2}^2}(\\mathbf{x}_\\mathbf{A})\n",
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"&& \\dots\n",
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"&& \\dfrac{\\partial^2 f}{\\partial x_2\\, \\partial x_n}(\\mathbf{x}_\\mathbf{A})\\\\\n",
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"\\vdots && \\vdots && \\ddots && \\vdots \\\\\n",
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"& \\dfrac{\\partial^2 f}{\\partial {x_2}^2}(\\mathbf{x}_\\mathbf{A})\n",
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"& \\dots\n",
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"& \\dfrac{\\partial^2 f}{\\partial x_2\\, \\partial x_n}(\\mathbf{x}_\\mathbf{A})\\\\\n",
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"\\vdots & \\vdots & \\ddots & \\vdots \\\\\n",
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"\\dfrac{\\partial^2 f}{\\partial x_n\\,\\partial x_1}(\\mathbf{x}_\\mathbf{A})\n",
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"&& \\dfrac{\\partial^2 f}{\\partial x_n\\,\\partial x_2}(\\mathbf{x}_\\mathbf{A})\n",
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"&& \\dots\n",
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"&& \\dfrac{\\partial^2 f}{\\partial {x_n}^2}(\\mathbf{x}_\\mathbf{A})\\\\\n",
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"& \\dfrac{\\partial^2 f}{\\partial x_n\\,\\partial x_2}(\\mathbf{x}_\\mathbf{A})\n",
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"& \\dots\n",
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"& \\dfrac{\\partial^2 f}{\\partial {x_n}^2}(\\mathbf{x}_\\mathbf{A})\\\\\n",
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"\\end{pmatrix}\n",
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"$"
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]
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@ -1305,12 +1305,12 @@
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"## Constant: $f(x)=c$\n",
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"\n",
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"$\n",
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"\\begin{split}\n",
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"f'(x) && = \\underset{\\epsilon \\to 0}\\lim\\dfrac{f(x+\\epsilon) - f(x)}{\\epsilon} && \\quad\\text{by definition}\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{c - c}{\\epsilon} && \\quad \\text{using }f(x) = c \\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim 0 && \\quad \\text{since }c - c = 0\\\\\n",
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"&& = 0 && \\quad \\text{since the limit of a constant is that constant}\n",
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"\\end{split}\n",
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"\\begin{align*}\n",
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"f'(x) & = \\underset{\\epsilon \\to 0}\\lim\\dfrac{f(x+\\epsilon) - f(x)}{\\epsilon} & \\quad\\text{by definition}\\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{c - c}{\\epsilon} & \\quad \\text{using }f(x) = c \\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim 0 & \\quad \\text{since }c - c = 0\\\\\n",
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"& = 0 & \\quad \\text{since the limit of a constant is that constant}\n",
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"\\end{align*}\n",
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"$\n"
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]
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},
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@ -1324,18 +1324,18 @@
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"## Product rule: $f(x)=g(x)h(x)$\n",
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"\n",
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"$\n",
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"\\begin{split}\n",
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"f'(x) && = \\underset{\\epsilon \\to 0}\\lim\\dfrac{f(x+\\epsilon) - f(x)}{\\epsilon} && \\quad\\text{by definition}\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{g(x+\\epsilon)h(x+\\epsilon) - g(x)h(x)}{\\epsilon} && \\quad \\text{using }f(x) = g(x)h(x) \\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{g(x+\\epsilon)h(x+\\epsilon) - g(x)h(x+\\epsilon) + g(x)h(x + \\epsilon) - g(x)h(x)}{\\epsilon} && \\quad \\text{subtracting and adding }g(x)h(x + \\epsilon)\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{g(x+\\epsilon)h(x+\\epsilon) - g(x)h(x+\\epsilon)}{\\epsilon} + \\underset{\\epsilon \\to 0}\\lim\\dfrac{g(x)h(x + \\epsilon) - g(x)h(x)}{\\epsilon} && \\quad \\text{since the limit of a sum is the sum of the limits}\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{g(x+\\epsilon) - g(x)}{\\epsilon}h(x+\\epsilon)\\right]} \\,+\\, \\underset{\\epsilon \\to 0}\\lim{\\left[g(x)\\dfrac{h(x + \\epsilon) - h(x)}{\\epsilon}\\right]} && \\quad \\text{factorizing }h(x+\\epsilon) \\text{ and } g(x)\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{g(x+\\epsilon) - g(x)}{\\epsilon}h(x+\\epsilon)\\right]} \\,+\\, g(x)\\underset{\\epsilon \\to 0}\\lim{\\dfrac{h(x + \\epsilon) - h(x)}{\\epsilon}} && \\quad \\text{taking } g(x) \\text{ out of the limit since it does not depend on }\\epsilon\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{g(x+\\epsilon) - g(x)}{\\epsilon}h(x+\\epsilon)\\right]} \\,+\\, g(x)h'(x) && \\quad \\text{using the definition of h'(x)}\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{g(x+\\epsilon) - g(x)}{\\epsilon}\\right]}\\underset{\\epsilon \\to 0}\\lim{h(x+\\epsilon)} + g(x)h'(x) && \\quad \\text{since the limit of a product is the product of the limits}\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{g(x+\\epsilon) - g(x)}{\\epsilon}\\right]}h(x) + h(x)g'(x) && \\quad \\text{since } h(x) \\text{ is continuous}\\\\\n",
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"&& = g'(x)h(x) + g(x)h'(x) && \\quad \\text{using the definition of }g'(x)\n",
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"\\end{split}\n",
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"\\begin{align*}\n",
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"f'(x) & = \\underset{\\epsilon \\to 0}\\lim\\dfrac{f(x+\\epsilon) - f(x)}{\\epsilon} & \\quad\\text{by definition}\\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{g(x+\\epsilon)h(x+\\epsilon) - g(x)h(x)}{\\epsilon} & \\quad \\text{using }f(x) = g(x)h(x) \\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{g(x+\\epsilon)h(x+\\epsilon) - g(x)h(x+\\epsilon) + g(x)h(x + \\epsilon) - g(x)h(x)}{\\epsilon} & \\quad \\text{subtracting and adding }g(x)h(x + \\epsilon)\\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{g(x+\\epsilon)h(x+\\epsilon) - g(x)h(x+\\epsilon)}{\\epsilon} + \\underset{\\epsilon \\to 0}\\lim\\dfrac{g(x)h(x + \\epsilon) - g(x)h(x)}{\\epsilon} & \\quad \\text{since the limit of a sum is the sum of the limits}\\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{g(x+\\epsilon) - g(x)}{\\epsilon}h(x+\\epsilon)\\right]} \\,+\\, \\underset{\\epsilon \\to 0}\\lim{\\left[g(x)\\dfrac{h(x + \\epsilon) - h(x)}{\\epsilon}\\right]} & \\quad \\text{factorizing }h(x+\\epsilon) \\text{ and } g(x)\\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{g(x+\\epsilon) - g(x)}{\\epsilon}h(x+\\epsilon)\\right]} \\,+\\, g(x)\\underset{\\epsilon \\to 0}\\lim{\\dfrac{h(x + \\epsilon) - h(x)}{\\epsilon}} & \\quad \\text{taking } g(x) \\text{ out of the limit since it does not depend on }\\epsilon\\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{g(x+\\epsilon) - g(x)}{\\epsilon}h(x+\\epsilon)\\right]} \\,+\\, g(x)h'(x) & \\quad \\text{using the definition of h'(x)}\\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{g(x+\\epsilon) - g(x)}{\\epsilon}\\right]}\\underset{\\epsilon \\to 0}\\lim{h(x+\\epsilon)} + g(x)h'(x) & \\quad \\text{since the limit of a product is the product of the limits}\\\\\n",
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"& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{g(x+\\epsilon) - g(x)}{\\epsilon}\\right]}h(x) + h(x)g'(x) & \\quad \\text{since } h(x) \\text{ is continuous}\\\\\n",
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"& = g'(x)h(x) + g(x)h'(x) & \\quad \\text{using the definition of }g'(x)\n",
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"\\end{align*}\n",
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"$\n",
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"\n",
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"Note that if $g(x)=c$ (a constant), then $g'(x)=0$, so the equation simplifies to:\n",
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"## Chain rule: $f(x)=g(h(x))$\n",
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"\n",
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"$\n",
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"\\begin{split}\n",
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"f'(x) && = \\underset{\\epsilon \\to 0}\\lim\\dfrac{f(x+\\epsilon) - f(x)}{\\epsilon} && \\quad\\text{by definition}\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{g(h(x+\\epsilon)) - g(h(x))}{\\epsilon} && \\quad \\text{using }f(x) = g(h(x))\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{h(x+\\epsilon)-h(x)}{h(x+\\epsilon)-h(x)}\\,\\dfrac{g(h(x+\\epsilon)) - g(h(x))}{\\epsilon}\\right]} && \\quad \\text{multiplying and dividing by }h(x+\\epsilon) - h(x)\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{h(x+\\epsilon)-h(x)}{\\epsilon}\\,\\dfrac{g(h(x+\\epsilon)) - g(h(x))}{h(x+\\epsilon)-h(x)}\\right]} && \\quad \\text{swapping the denominators}\\\\\n",
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"&& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{h(x+\\epsilon)-h(x)}{\\epsilon}\\right]} \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{g(h(x+\\epsilon)) - g(h(x))}{h(x+\\epsilon)-h(x)}\\right]} && \\quad \\text{the limit of a product is the product of the limits}\\\\\n",
|
||||
"&& = h'(x) \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{g(h(x+\\epsilon)) - g(h(x))}{h(x+\\epsilon)-h(x)}\\right]} && \\quad \\text{using the definition of }h'(x)\\\\\n",
|
||||
"&& = h'(x) \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{g(u) - g(v)}{u-v}\\right]} && \\quad \\text{using }u=h(x+\\epsilon) \\text{ and } v=h(x)\\\\\n",
|
||||
"&& = h'(x) \\underset{u \\to v}\\lim{\\left[\\dfrac{g(u) - g(v)}{u-v}\\right]} && \\quad \\text{ since } h \\text{ is continuous, so } \\underset{\\epsilon \\to 0}\\lim{u}=v\\\\\n",
|
||||
"&& = h'(x)g'(v) && \\quad \\text{ using the definition of } g'(v)\\\\\n",
|
||||
"&& = h'(x)g'(h(x)) && \\quad \\text{ since } v = h(x)\n",
|
||||
"\\end{split}\n",
|
||||
"\\begin{align*}\n",
|
||||
"f'(x) & = \\underset{\\epsilon \\to 0}\\lim\\dfrac{f(x+\\epsilon) - f(x)}{\\epsilon} & \\quad\\text{by definition}\\\\\n",
|
||||
"& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{g(h(x+\\epsilon)) - g(h(x))}{\\epsilon} & \\quad \\text{using }f(x) = g(h(x))\\\\\n",
|
||||
"& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{h(x+\\epsilon)-h(x)}{h(x+\\epsilon)-h(x)}\\,\\dfrac{g(h(x+\\epsilon)) - g(h(x))}{\\epsilon}\\right]} & \\quad \\text{multiplying and dividing by }h(x+\\epsilon) - h(x)\\\\\n",
|
||||
"& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{h(x+\\epsilon)-h(x)}{\\epsilon}\\,\\dfrac{g(h(x+\\epsilon)) - g(h(x))}{h(x+\\epsilon)-h(x)}\\right]} & \\quad \\text{swapping the denominators}\\\\\n",
|
||||
"& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{h(x+\\epsilon)-h(x)}{\\epsilon}\\right]} \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{g(h(x+\\epsilon)) - g(h(x))}{h(x+\\epsilon)-h(x)}\\right]} & \\quad \\text{the limit of a product is the product of the limits}\\\\\n",
|
||||
"& = h'(x) \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{g(h(x+\\epsilon)) - g(h(x))}{h(x+\\epsilon)-h(x)}\\right]} & \\quad \\text{using the definition of }h'(x)\\\\\n",
|
||||
"& = h'(x) \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{g(u) - g(v)}{u-v}\\right]} & \\quad \\text{using }u=h(x+\\epsilon) \\text{ and } v=h(x)\\\\\n",
|
||||
"& = h'(x) \\underset{u \\to v}\\lim{\\left[\\dfrac{g(u) - g(v)}{u-v}\\right]} & \\quad \\text{ since } h \\text{ is continuous, so } \\underset{\\epsilon \\to 0}\\lim{u}=v\\\\\n",
|
||||
"& = h'(x)g'(v) & \\quad \\text{ using the definition of } g'(v)\\\\\n",
|
||||
"& = h'(x)g'(h(x)) & \\quad \\text{ since } v = h(x)\n",
|
||||
"\\end{align*}\n",
|
||||
"$"
|
||||
]
|
||||
},
|
||||
|
|
@ -1380,15 +1380,15 @@
|
|||
"There are several equivalent definitions of the number $e$. One of them states that $e$ is the unique positive number for which $\\underset{\\epsilon \\to 0}\\lim{\\dfrac{e^\\epsilon - 1}{\\epsilon}}=1$. We will use this in this proof:\n",
|
||||
"\n",
|
||||
"$\n",
|
||||
"\\begin{split}\n",
|
||||
"f'(x) && = \\underset{\\epsilon \\to 0}\\lim\\dfrac{f(x+\\epsilon) - f(x)}{\\epsilon} && \\quad\\text{by definition}\\\\\n",
|
||||
"&& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{e^{x+\\epsilon} - e^x}{\\epsilon} && \\quad \\text{using }f(x) = e^x\\\\\n",
|
||||
"&& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{e^x e^\\epsilon - e^x}{\\epsilon} && \\quad \\text{using the fact that } x^{a+b}=x^a x^b\\\\\n",
|
||||
"&& = \\underset{\\epsilon \\to 0}\\lim{\\left[e^x\\dfrac{e^\\epsilon - 1}{\\epsilon}\\right]} && \\quad \\text{factoring out }e^x\\\\\n",
|
||||
"&& = \\underset{\\epsilon \\to 0}\\lim{e^x} \\, \\underset{\\epsilon \\to 0}\\lim{\\dfrac{e^\\epsilon - 1}{\\epsilon}} && \\quad \\text{the limit of a product is the product of the limits}\\\\\n",
|
||||
"&& = \\underset{\\epsilon \\to 0}\\lim{e^x} && \\quad \\text{since }\\underset{\\epsilon \\to 0}\\lim{\\dfrac{e^\\epsilon - 1}{\\epsilon}}=1\\\\\n",
|
||||
"&& = e^x && \\quad \\text{since } e^x \\text{ does not depend on }\\epsilon\n",
|
||||
"\\end{split}\n",
|
||||
"\\begin{align*}\n",
|
||||
"f'(x) & = \\underset{\\epsilon \\to 0}\\lim\\dfrac{f(x+\\epsilon) - f(x)}{\\epsilon} & \\quad\\text{by definition}\\\\\n",
|
||||
"& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{e^{x+\\epsilon} - e^x}{\\epsilon} & \\quad \\text{using }f(x) = e^x\\\\\n",
|
||||
"& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{e^x e^\\epsilon - e^x}{\\epsilon} & \\quad \\text{using the fact that } x^{a+b}=x^a x^b\\\\\n",
|
||||
"& = \\underset{\\epsilon \\to 0}\\lim{\\left[e^x\\dfrac{e^\\epsilon - 1}{\\epsilon}\\right]} & \\quad \\text{factoring out }e^x\\\\\n",
|
||||
"& = \\underset{\\epsilon \\to 0}\\lim{e^x} \\, \\underset{\\epsilon \\to 0}\\lim{\\dfrac{e^\\epsilon - 1}{\\epsilon}} & \\quad \\text{the limit of a product is the product of the limits}\\\\\n",
|
||||
"& = \\underset{\\epsilon \\to 0}\\lim{e^x} & \\quad \\text{since }\\underset{\\epsilon \\to 0}\\lim{\\dfrac{e^\\epsilon - 1}{\\epsilon}}=1\\\\\n",
|
||||
"& = e^x & \\quad \\text{since } e^x \\text{ does not depend on }\\epsilon\n",
|
||||
"\\end{align*}\n",
|
||||
"$\n"
|
||||
]
|
||||
},
|
||||
|
|
@ -1412,19 +1412,19 @@
|
|||
"This will come in handy in a second:\n",
|
||||
"\n",
|
||||
"$\n",
|
||||
"\\begin{split}\n",
|
||||
"f'(x) && = \\underset{\\epsilon \\to 0}\\lim\\dfrac{f(x+\\epsilon) - f(x)}{\\epsilon} && \\quad\\text{by definition}\\\\\n",
|
||||
"&& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{\\ln(x+\\epsilon) - \\ln(x)}{\\epsilon} && \\quad \\text{using }f(x) = \\ln(x)\\\\\n",
|
||||
"&& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{\\ln\\left(\\dfrac{x+\\epsilon}{x}\\right)}{\\epsilon} && \\quad \\text{since }\\ln(a)-\\ln(b)=\\ln\\left(\\dfrac{a}{b}\\right)\\\\\n",
|
||||
"&& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{1}{\\epsilon} \\, \\ln\\left(1 + \\dfrac{\\epsilon}{x}\\right)\\right]} && \\quad \\text{just moving things around a bit}\\\\\n",
|
||||
"&& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{1}{xu} \\, \\ln\\left(1 + u\\right)\\right]} && \\quad \\text{defining }u=\\dfrac{\\epsilon}{x} \\text{ and thus } \\epsilon=xu\\\\\n",
|
||||
"&& = \\underset{u \\to 0}\\lim{\\left[\\dfrac{1}{xu} \\, \\ln\\left(1 + u\\right)\\right]} && \\quad \\text{replacing } \\underset{\\epsilon \\to 0}\\lim \\text{ with } \\underset{u \\to 0}\\lim \\text{ since }\\underset{\\epsilon \\to 0}\\lim u=0\\\\\n",
|
||||
"&& = \\underset{u \\to 0}\\lim{\\left[\\dfrac{1}{x} \\, \\ln\\left((1 + u)^{1/u}\\right)\\right]} && \\quad \\text{since }a\\ln(b)=\\ln(a^b)\\\\\n",
|
||||
"&& = \\dfrac{1}{x}\\underset{u \\to 0}\\lim{\\left[\\ln\\left((1 + u)^{1/u}\\right)\\right]} && \\quad \\text{taking }\\dfrac{1}{x} \\text{ out since it does not depend on }\\epsilon\\\\\n",
|
||||
"&& = \\dfrac{1}{x}\\ln\\left(\\underset{u \\to 0}\\lim{(1 + u)^{1/u}}\\right) && \\quad \\text{taking }\\ln\\text{ out since it is a continuous function}\\\\\n",
|
||||
"&& = \\dfrac{1}{x}\\ln(e) && \\quad \\text{since }e=\\underset{u \\to 0}\\lim{(1 + u)^{1/u}}\\\\\n",
|
||||
"&& = \\dfrac{1}{x} && \\quad \\text{since }\\ln(e)=1\n",
|
||||
"\\end{split}\n",
|
||||
"\\begin{align*}\n",
|
||||
"f'(x) & = \\underset{\\epsilon \\to 0}\\lim\\dfrac{f(x+\\epsilon) - f(x)}{\\epsilon} & \\quad\\text{by definition}\\\\\n",
|
||||
"& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{\\ln(x+\\epsilon) - \\ln(x)}{\\epsilon} & \\quad \\text{using }f(x) = \\ln(x)\\\\\n",
|
||||
"& = \\underset{\\epsilon \\to 0}\\lim\\dfrac{\\ln\\left(\\dfrac{x+\\epsilon}{x}\\right)}{\\epsilon} & \\quad \\text{since }\\ln(a)-\\ln(b)=\\ln\\left(\\dfrac{a}{b}\\right)\\\\\n",
|
||||
"& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{1}{\\epsilon} \\, \\ln\\left(1 + \\dfrac{\\epsilon}{x}\\right)\\right]} & \\quad \\text{just moving things around a bit}\\\\\n",
|
||||
"& = \\underset{\\epsilon \\to 0}\\lim{\\left[\\dfrac{1}{xu} \\, \\ln\\left(1 + u\\right)\\right]} & \\quad \\text{defining }u=\\dfrac{\\epsilon}{x} \\text{ and thus } \\epsilon=xu\\\\\n",
|
||||
"& = \\underset{u \\to 0}\\lim{\\left[\\dfrac{1}{xu} \\, \\ln\\left(1 + u\\right)\\right]} & \\quad \\text{replacing } \\underset{\\epsilon \\to 0}\\lim \\text{ with } \\underset{u \\to 0}\\lim \\text{ since }\\underset{\\epsilon \\to 0}\\lim u=0\\\\\n",
|
||||
"& = \\underset{u \\to 0}\\lim{\\left[\\dfrac{1}{x} \\, \\ln\\left((1 + u)^{1/u}\\right)\\right]} & \\quad \\text{since }a\\ln(b)=\\ln(a^b)\\\\\n",
|
||||
"& = \\dfrac{1}{x}\\underset{u \\to 0}\\lim{\\left[\\ln\\left((1 + u)^{1/u}\\right)\\right]} & \\quad \\text{taking }\\dfrac{1}{x} \\text{ out since it does not depend on }\\epsilon\\\\\n",
|
||||
"& = \\dfrac{1}{x}\\ln\\left(\\underset{u \\to 0}\\lim{(1 + u)^{1/u}}\\right) & \\quad \\text{taking }\\ln\\text{ out since it is a continuous function}\\\\\n",
|
||||
"& = \\dfrac{1}{x}\\ln(e) & \\quad \\text{since }e=\\underset{u \\to 0}\\lim{(1 + u)^{1/u}}\\\\\n",
|
||||
"& = \\dfrac{1}{x} & \\quad \\text{since }\\ln(e)=1\n",
|
||||
"\\end{align*}\n",
|
||||
"$\n"
|
||||
]
|
||||
},
|
||||
|
|
@ -1655,16 +1655,16 @@
|
|||
"Now the second thing we need to prove before we can tackle the derivative of the $\\sin$ function is the fact that $\\underset{\\theta \\to 0}\\lim\\dfrac{\\cos(\\theta) - 1}{\\theta}=0$. Here we go:\n",
|
||||
"\n",
|
||||
"$\n",
|
||||
"\\begin{split}\n",
|
||||
"\\underset{\\theta \\to 0}\\lim\\dfrac{\\cos(\\theta) - 1}{\\theta} && = \\underset{\\theta \\to 0}\\lim\\dfrac{\\cos(\\theta) - 1}{\\theta}\\frac{\\cos(\\theta) + 1}{\\cos(\\theta) + 1} && \\quad \\text{ multiplying and dividing by }\\cos(\\theta)+1\\\\\n",
|
||||
"&& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\cos^2(\\theta) - 1}{\\theta(\\cos(\\theta) + 1)} && \\quad \\text{ since }(a-1)(a+1)=a^2-1\\\\\n",
|
||||
"&& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin^2(\\theta)}{\\theta(\\cos(\\theta) + 1)} && \\quad \\text{ since }\\cos^2(\\theta) - 1 = \\sin^2(\\theta)\\\\\n",
|
||||
"&& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(\\theta)}{\\theta}\\dfrac{\\sin(\\theta)}{\\cos(\\theta) + 1} && \\quad \\text{ just rearranging the terms}\\\\\n",
|
||||
"&& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(\\theta)}{\\theta} \\, \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(\\theta)}{\\cos(\\theta) + 1} && \\quad \\text{ since the limit of a product is the product of the limits}\\\\\n",
|
||||
"&& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(\\theta)}{\\cos(\\theta) + 1} && \\quad \\text{ since } \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(\\theta)}{\\theta}=1\\\\\n",
|
||||
"&& = \\dfrac{0}{1+1} && \\quad \\text{ since } \\underset{\\theta \\to 0}\\lim\\sin(\\theta)=0 \\text{ and } \\underset{\\theta \\to 0}\\lim\\cos(\\theta)=1\\\\\n",
|
||||
"&& = 0\\\\\n",
|
||||
"\\end{split}\n",
|
||||
"\\begin{align*}\n",
|
||||
"\\underset{\\theta \\to 0}\\lim\\dfrac{\\cos(\\theta) - 1}{\\theta} & = \\underset{\\theta \\to 0}\\lim\\dfrac{\\cos(\\theta) - 1}{\\theta}\\frac{\\cos(\\theta) + 1}{\\cos(\\theta) + 1} & \\quad \\text{ multiplying and dividing by }\\cos(\\theta)+1\\\\\n",
|
||||
"& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\cos^2(\\theta) - 1}{\\theta(\\cos(\\theta) + 1)} & \\quad \\text{ since }(a-1)(a+1)=a^2-1\\\\\n",
|
||||
"& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin^2(\\theta)}{\\theta(\\cos(\\theta) + 1)} & \\quad \\text{ since }\\cos^2(\\theta) - 1 = \\sin^2(\\theta)\\\\\n",
|
||||
"& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(\\theta)}{\\theta}\\dfrac{\\sin(\\theta)}{\\cos(\\theta) + 1} & \\quad \\text{ just rearranging the terms}\\\\\n",
|
||||
"& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(\\theta)}{\\theta} \\, \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(\\theta)}{\\cos(\\theta) + 1} & \\quad \\text{ since the limit of a product is the product of the limits}\\\\\n",
|
||||
"& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(\\theta)}{\\cos(\\theta) + 1} & \\quad \\text{ since } \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(\\theta)}{\\theta}=1\\\\\n",
|
||||
"& = \\dfrac{0}{1+1} & \\quad \\text{ since } \\underset{\\theta \\to 0}\\lim\\sin(\\theta)=0 \\text{ and } \\underset{\\theta \\to 0}\\lim\\cos(\\theta)=1\\\\\n",
|
||||
"& = 0\\\\\n",
|
||||
"\\end{align*}\n",
|
||||
"$\n",
|
||||
"\n",
|
||||
"<hr />\n",
|
||||
|
|
@ -1694,15 +1694,15 @@
|
|||
},
|
||||
"source": [
|
||||
"$\n",
|
||||
"\\begin{split}\n",
|
||||
"f'(x) && = \\underset{\\theta \\to 0}\\lim\\dfrac{f(x+\\theta) - f(x)}{\\theta} && \\quad\\text{by definition}\\\\\n",
|
||||
"&& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(x+\\theta) - \\sin(x)}{\\theta} && \\quad \\text{using }f(x) = \\sin(x)\\\\\n",
|
||||
"&& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\cos(x)\\sin(\\theta) + \\sin(x)\\cos(\\theta) - \\sin(x)}{\\theta} && \\quad \\text{since } cos(a+b)=\\cos(a)\\sin(b)+\\sin(a)\\cos(b)\\\\\n",
|
||||
"&& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\cos(x)\\sin(\\theta)}{\\theta} + \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(x)\\cos(\\theta) - \\sin(x)}{\\theta} && \\quad \\text{since the limit of a sum is the sum of the limits}\\\\\n",
|
||||
"&& = \\cos(x)\\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(\\theta)}{\\theta} + \\sin(x)\\underset{\\theta \\to 0}\\lim\\dfrac{\\cos(\\theta) - 1}{\\theta} && \\quad \\text{bringing out } \\cos(x) \\text{ and } \\sin(x) \\text{ since they don't depend on }\\theta\\\\\n",
|
||||
"&& = \\cos(x)\\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(\\theta)}{\\theta} && \\quad \\text{since }\\underset{\\theta \\to 0}\\lim\\dfrac{\\cos(\\theta) - 1}{\\theta}=0\\\\\n",
|
||||
"&& = \\cos(x) && \\quad \\text{since }\\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(\\theta)}{\\theta}=1\\\\\n",
|
||||
"\\end{split}\n",
|
||||
"\\begin{align*}\n",
|
||||
"f'(x) & = \\underset{\\theta \\to 0}\\lim\\dfrac{f(x+\\theta) - f(x)}{\\theta} & \\quad\\text{by definition}\\\\\n",
|
||||
"& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(x+\\theta) - \\sin(x)}{\\theta} & \\quad \\text{using }f(x) = \\sin(x)\\\\\n",
|
||||
"& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\cos(x)\\sin(\\theta) + \\sin(x)\\cos(\\theta) - \\sin(x)}{\\theta} & \\quad \\text{since } cos(a+b)=\\cos(a)\\sin(b)+\\sin(a)\\cos(b)\\\\\n",
|
||||
"& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\cos(x)\\sin(\\theta)}{\\theta} + \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(x)\\cos(\\theta) - \\sin(x)}{\\theta} & \\quad \\text{since the limit of a sum is the sum of the limits}\\\\\n",
|
||||
"& = \\cos(x)\\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(\\theta)}{\\theta} + \\sin(x)\\underset{\\theta \\to 0}\\lim\\dfrac{\\cos(\\theta) - 1}{\\theta} & \\quad \\text{bringing out } \\cos(x) \\text{ and } \\sin(x) \\text{ since they don't depend on }\\theta\\\\\n",
|
||||
"& = \\cos(x)\\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(\\theta)}{\\theta} & \\quad \\text{since }\\underset{\\theta \\to 0}\\lim\\dfrac{\\cos(\\theta) - 1}{\\theta}=0\\\\\n",
|
||||
"& = \\cos(x) & \\quad \\text{since }\\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(\\theta)}{\\theta}=1\\\\\n",
|
||||
"\\end{align*}\n",
|
||||
"$\n"
|
||||
]
|
||||
},
|
||||
|
|
@ -1718,16 +1718,16 @@
|
|||
"Since we have proven that $\\sin'(x)=\\cos(x)$, proving that $\\cos'(x)=-\\sin(x)$ will be much easier.\n",
|
||||
"\n",
|
||||
"$\n",
|
||||
"\\begin{split}\n",
|
||||
"f'(x) && = \\underset{\\theta \\to 0}\\lim\\dfrac{f(x+\\theta) - f(x)}{\\theta} && \\quad\\text{by definition}\\\\\n",
|
||||
"&& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\cos(x+\\theta) - \\cos(x)}{\\theta} && \\quad \\text{using }f(x) = \\cos(x)\\\\\n",
|
||||
"&& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin\\left(x+\\dfrac{\\pi}{2}+\\theta\\right) - \\sin\\left(x+\\dfrac{\\pi}{2}\\right)}{\\theta} && \\quad \\text{since }\\cos(x) = \\sin\\left(x+\\dfrac{\\pi}{2}\\right)\\\\\n",
|
||||
"&& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(z+\\theta) - \\sin(z)}{\\theta} && \\quad \\text{using }z = x + \\dfrac{\\pi}{2}\\\\\n",
|
||||
"&& = \\sin'(z) && \\quad \\text{using the definition of }\\sin'(z)\\\\\n",
|
||||
"&& = \\cos(z) && \\quad \\text{since we proved that }\\sin'(z)=\\cos(z)\\\\\n",
|
||||
"&& = \\cos\\left(x + \\dfrac{\\pi}{2}\\right) && \\quad \\text{using the definition of }z\\\\\n",
|
||||
"&& = -\\sin(x) && \\quad \\text{using this well-known rule of trigonometry}\n",
|
||||
"\\end{split}\n",
|
||||
"\\begin{align*}\n",
|
||||
"f'(x) & = \\underset{\\theta \\to 0}\\lim\\dfrac{f(x+\\theta) - f(x)}{\\theta} & \\quad\\text{by definition}\\\\\n",
|
||||
"& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\cos(x+\\theta) - \\cos(x)}{\\theta} & \\quad \\text{using }f(x) = \\cos(x)\\\\\n",
|
||||
"& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin\\left(x+\\dfrac{\\pi}{2}+\\theta\\right) - \\sin\\left(x+\\dfrac{\\pi}{2}\\right)}{\\theta} & \\quad \\text{since }\\cos(x) = \\sin\\left(x+\\dfrac{\\pi}{2}\\right)\\\\\n",
|
||||
"& = \\underset{\\theta \\to 0}\\lim\\dfrac{\\sin(z+\\theta) - \\sin(z)}{\\theta} & \\quad \\text{using }z = x + \\dfrac{\\pi}{2}\\\\\n",
|
||||
"& = \\sin'(z) & \\quad \\text{using the definition of }\\sin'(z)\\\\\n",
|
||||
"& = \\cos(z) & \\quad \\text{since we proved that }\\sin'(z)=\\cos(z)\\\\\n",
|
||||
"& = \\cos\\left(x + \\dfrac{\\pi}{2}\\right) & \\quad \\text{using the definition of }z\\\\\n",
|
||||
"& = -\\sin(x) & \\quad \\text{using this well-known rule of trigonometry}\n",
|
||||
"\\end{align*}\n",
|
||||
"$\n"
|
||||
]
|
||||
},
|
||||
|
|
|
|||
Loading…
Reference in New Issue