align fix; theorem style; condition number
This commit is contained in:
jverzani
@@ -22,6 +22,8 @@ nothing
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---
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{width=60%}
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Before defining the derivative of a function, let's begin with two motivating examples.
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@@ -520,13 +522,14 @@ This holds two rules: the derivative of a constant times a function is the const
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This example shows a useful template:
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$$
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\begin{align*}
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[2x^2 - \frac{x}{3} + 3e^x]' & = 2[\square]' - \frac{[\square]'}{3} + 3[\square]'\\
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&= 2[x^2]' - \frac{[x]'}{3} + 3[e^x]'\\
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&= 2(2x) - \frac{1}{3} + 3e^x\\
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&= 4x - \frac{1}{3} + 3e^x
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\end{align*}
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$$
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### Product rule
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@@ -548,12 +551,13 @@ The output uses the Leibniz notation to represent that the derivative of $u(x) \
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This example shows a useful template for the product rule:
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$$
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\begin{align*}
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[(x^2+1)\cdot e^x]' &= [\square]' \cdot (\square) + (\square) \cdot [\square]'\\
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&= [x^2 + 1]' \cdot (e^x) + (x^2+1) \cdot [e^x]'\\
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&= (2x)\cdot e^x + (x^2+1)\cdot e^x
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\end{align*}
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$$
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### Quotient rule
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@@ -572,12 +576,13 @@ limit((f(x+h) - f(x))/h, h => 0)
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This example shows a useful template for the quotient rule:
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$$
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\begin{align*}
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[\frac{x^2+1}{e^x}]' &= \frac{[\square]' \cdot (\square) - (\square) \cdot [\square]'}{(\square)^2}\\
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&= \frac{[x^2 + 1]' \cdot (e^x) - (x^2+1) \cdot [e^x]'}{(e^x)^2}\\
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&= \frac{(2x)\cdot e^x - (x^2+1)\cdot e^x}{e^{2x}}
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\end{align*}
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$$
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##### Examples
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@@ -672,19 +677,21 @@ There are $n$ terms, each where one of the $f_i$s have a derivative. Were we to
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With this, we can proceed. Each term $x-i$ has derivative $1$, so the answer to $f'(x)$, with $f$ as above, is
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$$
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\begin{align*}
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f'(x) &= f(x)/(x-1) + f(x)/(x-2) + f(x)/(x-3)\\
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&+ f(x)/(x-4) + f(x)/(x-5),
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\end{align*}
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$$
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That is
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$$
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\begin{align*}
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f'(x) &= (x-2)(x-3)(x-4)(x-5) + (x-1)(x-3)(x-4)(x-5)\\
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&+ (x-1)(x-2)(x-4)(x-5) + (x-1)(x-2)(x-3)(x-5) \\
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&+ (x-1)(x-2)(x-3)(x-4).
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\end{align*}
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$$
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---
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@@ -749,17 +756,18 @@ Combined, we would end up with:
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To see that this works in our specific case, we assume the general power rule that $[x^n]' = n x^{n-1}$ to get:
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$$
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\begin{align*}
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f(x) &= x^2 & g(x) &= \sqrt{x}\\
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f'(\square) &= 2(\square) & g'(x) &= \frac{1}{2}x^{-1/2}
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\end{align*}
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$$
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We use $\square$ for the argument of `f'` to emphasize that $g(x)$ is the needed value, not just $x$:
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$$
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\begin{align*}
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[(\sqrt{x})^2]' &= [f(g(x)]'\\
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&= f'(g(x)) \cdot g'(x) \\
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@@ -767,6 +775,7 @@ We use $\square$ for the argument of `f'` to emphasize that $g(x)$ is the needed
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&= \frac{2\sqrt{x}}{2\sqrt{x}}\\
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&=1
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\end{align*}
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$$
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This is the same as the derivative of $x$ found by first evaluating the composition. For this problem, the chain rule is not necessary, but typically it is a needed rule to fully differentiate a function.
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@@ -778,11 +787,12 @@ This is the same as the derivative of $x$ found by first evaluating the composit
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Find the derivative of $f(x) = \sqrt{1 - x^2}$. We identify the composition of $\sqrt{x}$ and $(1-x^2)$. We set the functions and their derivatives into a pattern to emphasize the pieces in the chain-rule formula:
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$$
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\begin{align*}
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f(x) &=\sqrt{x} = x^{1/2} & g(x) &= 1 - x^2 \\
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f'(\square) &=(1/2)(\square)^{-1/2} & g'(x) &= -2x
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\end{align*}
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$$
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Then:
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@@ -823,11 +833,12 @@ This is a useful rule to remember for expressions involving exponentials.
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Find the derivative of $\sin(x)\cos(2x)$ at $x=\pi$.
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$$
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\begin{align*}
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[\sin(x)\cos(2x)]'\big|_{x=\pi} &=(\cos(x)\cos(2x) + \sin(x)(-\sin(2x)\cdot 2))\big|_{x=\pi} \\
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& =((-1)(1) + (0)(-0)(2)) = -1.
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\end{align*}
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$$
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##### Proof of the Chain Rule
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@@ -844,23 +855,25 @@ g(a+h) = g(a) + g'(a)h + \epsilon_g(h) h = g(a) + h',
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$$
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Where $h' = (g'(a) + \epsilon_g(h))h \rightarrow 0$ as $h \rightarrow 0$ will be used to simplify the following:
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$$
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\begin{align*}
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f(g(a+h)) - f(g(a)) &=
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f(g(a) + g'(a)h + \epsilon_g(h)h) - f(g(a)) \\
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&= f(g(a)) + f'(g(a)) (g'(a)h + \epsilon_g(h)h) + \epsilon_f(h')(h') - f(g(a))\\
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&= f'(g(a)) g'(a)h + f'(g(a))(\epsilon_g(h)h) + \epsilon_f(h')(h').
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\end{align*}
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$$
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Rearranging:
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$$
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\begin{align*}
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f(g(a+h)) &- f(g(a)) - f'(g(a)) g'(a) h\\
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&= f'(g(a))\epsilon_g(h)h + \epsilon_f(h')(h')\\
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&=(f'(g(a)) \epsilon_g(h) + \epsilon_f(h') (g'(a) + \epsilon_g(h)))h \\
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&=\epsilon(h)h,
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\end{align*}
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$$
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where $\epsilon(h)$ combines the above terms which go to zero as $h\rightarrow 0$ into one. This is the alternative definition of the derivative, showing $(f\circ g)'(a) = f'(g(a)) g'(a)$ when $g$ is differentiable at $a$ and $f$ is differentiable at $g(a)$.
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@@ -871,17 +884,18 @@ where $\epsilon(h)$ combines the above terms which go to zero as $h\rightarrow 0
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The chain rule name could also be simply the "composition rule," as that is the operation the rule works for. However, in practice, there are usually *multiple* compositions, and the "chain" rule is used to chain together the different pieces. To get a sense, consider a triple composition $u(v(w(x)))$. This will have derivative:
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$$
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\begin{align*}
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[u(v(w(x)))]' &= u'(v(w(x))) \cdot [v(w(x))]' \\
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&= u'(v(w(x))) \cdot v'(w(x)) \cdot w'(x)
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\end{align*}
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$$
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The answer can be viewed as a repeated peeling off of the outer function, a view with immediate application to many compositions. To see that in action with an expression, consider this derivative problem, shown in steps:
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$$
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\begin{align*}
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[\sin(e^{\cos(x^2-x)})]'
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&= \cos(e^{\cos(x^2-x)}) \cdot [e^{\cos(x^2-x)}]'\\
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@@ -889,6 +903,7 @@ The answer can be viewed as a repeated peeling off of the outer function, a view
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&= \cos(e^{\cos(x^2-x)}) \cdot e^{\cos(x^2-x)} \cdot (-\sin(x^2-x)) \cdot [x^2-x]'\\
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&= \cos(e^{\cos(x^2-x)}) \cdot e^{\cos(x^2-x)} \cdot (-\sin(x^2-x)) \cdot (2x-1)\\
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\end{align*}
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$$
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##### More examples of differentiation
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@@ -1004,7 +1019,7 @@ Find the derivative of $f(x) = x \cdot e^{-x^2}$.
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Using the product rule and then the chain rule, we have:
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$$
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\begin{align*}
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f'(x) &= [x \cdot e^{-x^2}]'\\
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&= [x]' \cdot e^{-x^2} + x \cdot [e^{-x^2}]'\\
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@@ -1012,6 +1027,7 @@ f'(x) &= [x \cdot e^{-x^2}]'\\
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&= e^{-x^2} + x \cdot e^{-x^2} \cdot (-2x)\\
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&= e^{-x^2} (1 - 2x^2).
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\end{align*}
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$$
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---
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@@ -1022,7 +1038,7 @@ Find the derivative of $f(x) = e^{-ax} \cdot \sin(x)$.
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Using the product rule and then the chain rule, we have:
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$$
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\begin{align*}
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f'(x) &= [e^{-ax} \cdot \sin(x)]'\\
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&= [e^{-ax}]' \cdot \sin(x) + e^{-ax} \cdot [\sin(x)]'\\
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@@ -1030,6 +1046,7 @@ f'(x) &= [e^{-ax} \cdot \sin(x)]'\\
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&= e^{-ax} \cdot (-a) \cdot \sin(x) + e^{-ax} \cos(x)\\
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&= e^{-ax}(\cos(x) - a\sin(x)).
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\end{align*}
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$$
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---
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@@ -1164,13 +1181,14 @@ Find the first $3$ derivatives of $f(x) = ax^3 + bx^2 + cx + d$.
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Differentiating a polynomial is done with the sum rule, here we repeat three times:
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$$
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\begin{align*}
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f(x) &= ax^3 + bx^2 + cx + d\\
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f'(x) &= 3ax^2 + 2bx + c \\
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f''(x) &= 3\cdot 2 a x + 2b \\
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f'''(x) &= 6a
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\end{align*}
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$$
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We can see, the fourth derivative – and all higher order ones – would be identically $0$. This is part of a general phenomenon: an $n$th degree polynomial has only $n$ non-zero derivatives.
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@@ -1181,7 +1199,7 @@ We can see, the fourth derivative – and all higher order ones – would be ide
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Find the first $5$ derivatives of $\sin(x)$.
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$$
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\begin{align*}
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f(x) &= \sin(x) \\
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f'(x) &= \cos(x) \\
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@@ -1190,6 +1208,7 @@ f'''(x) &= -\cos(x) \\
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f^{(4)} &= \sin(x) \\
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f^{(5)} &= \cos(x)
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\end{align*}
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$$
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We see the derivatives repeat themselves. (We also see alternative notation for higher order derivatives.)
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@@ -1616,13 +1635,14 @@ The right graph is of $g(x) = \exp(x)$ at $x=1$, the left graph of $f(x) = \sin(
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Assuming the approximation gets better for $h$ close to $0$, as it visually does, the derivative at $1$ for $f(g(x))$ should be given by this limit:
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$$
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\begin{align*}
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\frac{d(f\circ g)}{dx}\mid_{x=1}
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&= \lim_{h\rightarrow 0} \frac{f(g(1) + g'(1)h)-f(g(1))}{h}\\
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&= \lim_{h\rightarrow 0} \frac{f(g(1) + g'(1)h)-f(g(1))}{g'(1)h} \cdot g'(1)\\
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&= \lim_{h\rightarrow 0} (f\circ g)'(g(1)) \cdot g'(1).
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\end{align*}
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$$
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What limit law, described below assuming all limits exist. allows the last equals sign?
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