합성함수의 미분법

$-12$

Applying the chain rule:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\{\cos(5x+3)\}&=-\sin( 5x+3)\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{ 5x+3\}\\ &=-\sin(5x+3)\cdot 5 \end{aligned}$$

Using the chain rule,

$$\begin{aligned} f'(x)&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{\left({x^2+2}\right)^5\right\}\\ &=5\left( x^2+2\right)^4\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{ x^2+2\}\\ &=5(x^2+2)^4 \cdot 2x\\ &=10x(x^2+2)^4 \end{aligned}$$

Using the chain rule,

$$\begin{aligned} T'(k)&=\frac{\mathrm{d}}{\mathrm{d}k}\left\{\left({4k^4+2k^2+1}\right)^{17}\right\}\\ &=17( 4k^4+2k^2+1)^{16}\cdot\frac{\mathrm{d}}{\mathrm{d}k}\{ 4k^4+2k^2+1\}\\ &=17(4k^4+2k^2+1)^{16}\cdot(16k^3+4k) \end{aligned}$$

Using the chain rule:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{\frac{x^2+1}{x^2-1}}\right\}&=\frac{1}{2\sqrt{ \frac{x^2+1}{x^2-1}}}\cdot \frac{\mathrm{d}}{\mathrm{d}x}\left\{ \frac{x^2+1}{x^2-1}\right\}\\ &=\frac{1}{2}\sqrt{\frac{x^2-1}{x^2+1}}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{x^2+1}{x^2-1}\right\} \\ \text{And now, the quotient rule:} \\ &=\frac{1}{2}\sqrt{\frac{x^2-1}{x^2+1}}\cdot\left(\frac{(x^2-1)(2x)-(x^2+1)2x}{(x^2-1)^2}\right)\\ &=\frac{1}{2}\sqrt{\frac{x^2-1}{x^2+1}}\cdot\left(\frac{-4x}{(x^2-1)^2}\right)\\ &=\sqrt{\frac{x^2-1}{x^2+1}}\cdot\left(\frac{-2x}{(x^2-1)^2}\right)\\ &=\frac{-2x}{({x^2-1})\sqrt{x^4-1}} \end{aligned}$$

If we let $g(x)=e^x$ and $h(x)=\cos(x^2)$, then $f(x)=g(h(x))$, so $f'(x)=g'(h(x))\cdot h'(x)$.

$$\begin{aligned} f'(x)&=e^{ \cos(x^2)}\cdot \frac{\mathrm{d}}{\mathrm{d}x}\{ \cos(x^2)\} \\ \text{In order to evaluate $\frac{\mathrm{d}}{\mathrm{d}x}\{{\cos(x^2)}\}$, we'll need the chain rule \textit{again}.} \\ &=e^{\cos(x^2)}\cdot [-\sin( x^2)]\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{ x^2\}\\ &=-e^{\cos(x^2)}\cdot \sin(x^2)\cdot 2x \end{aligned}$$

We use the chain rule, followed by the quotient rule:

$$\begin{aligned} f'(x) &=g'\left( \frac{x}{h(x)}\right)\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \frac{x}{h(x)} \right\} \\&= g'\left(\frac{x}{h(x)}\right)\cdot\frac{h(x)-xh'(x)}{h(x)^2} \\ \text{When $x=2$:} \\ f'(2) &= g'\left(\frac{2}{h(2)}\right)\frac{h(2)-2h'(2)}{h(2)^2}\\ &= 4\frac{2-2\times3}{2^2} =-4 \end{aligned}$$

Using the chain rule, followed by the product rule:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{x\cos(x)}\right\}&=e^{ x\cos x}\frac{\mathrm{d}}{\mathrm{d}x}\left\{ x\cos x\right\} \\&=[\cos x -x\sin x]e^{x\cos(x)} \end{aligned}$$

Using the chain rule:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{x^2+\cos(x)}\right\}&=e^{ x^2+\cos x}\frac{\mathrm{d}}{\mathrm{d}x}\left\{ x^2+\cos x\right\} \\&=[2x-\sin x]e^{x^2+\cos(x)} \end{aligned}$$

Using the chain rule, followed by the quotient rule:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{ \sqrt{\dfrac{x-1}{x+2}}\right\}&=\frac{1}{2\sqrt{ \dfrac{x-1}{x+2}}}\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \dfrac{x-1}{x+2}\right\} \\&=\frac{\sqrt{x+2}}{2\sqrt{x-1}}\cdot \frac{(x+2)-(x-1)}{(x+2)^2}\\ &=\frac{3}{2\sqrt{x-1}\sqrt{x+2}^3} \end{aligned}$$

First, we manipulate our function to make it easier to differentiate:

$$\begin{aligned} f(x)&=x^{-2}+(x^2-1)^{1/2} \\ \text{Now, we can use the power rule to differentiate $\dfrac{1}{x^2}$. This will be easier than differentiating $\dfrac{1}{x^2}$ using quotient rule, but if you prefer, quotient rule will also work.} \\ f'(x)&=-2x^{-3}+\frac{1}{2}( x^2-1)^{-1/2}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{ x^2-1\}\\ &=-2x^{-3}+\frac{1}{2}({x^2-1})^{-1/2}(2x)\\ &=\frac{-2}{x^3}+\frac{x}{\sqrt{x^2-1}} \end{aligned}$$

The function $f(x)$ is only defined when $x \neq 0$ and when $x^2-1 \geq 0$. That is, when $x$ is in $(-\infty,-1] \cup [1,\infty)$. We have an added restriction on the domain of $f'(x)$: $x^2-1$ must not be zero. So, the domain of $f'(x)$ is $(-\infty,-1)\cup(1,\infty)$.

We use the quotient rule, noting that $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\{\sin 5x\}=5\cos 5x$:

$$\begin{aligned} f'(x)&=\frac{(1+x^2)(5\cos 5x)-(\sin 5x)(2x)}{{(1+x^2)}^2} \end{aligned}$$

If we let $g(x)=\sec x$ and $h(x)=e^{2x+7}$, then $f(x)=g(h(x))$, so by the chain rule, $f'(x)=g'(h(x))\cdot h'(x)$. Since $g'(x)=\sec x \tan x$:

$$\begin{aligned} f'(x)&=g'(h(x))\cdot h'(x)\\ &=\sec(h(x))\tan(h(x)) \cdot h'(x)\\ &=\sec(e^{2x+7})\tan(e^{2x+7}) \cdot \frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{2x+7}\right\} \\ \text{Here, we need the chain rule again:} \\ &=\sec(e^{2x+7})\tan(e^{2x+7}) \cdot \left[e^{ 2x+7}\cdot \frac{\mathrm{d}}{\mathrm{d}x}\{ 2x+7\}\right]\\ &=\sec(e^{2x+7})\tan(e^{2x+7}) \cdot \left[e^{2x+7}\cdot2\right]\\ &=2e^{2x+7}\sec(e^{2x+7})\tan(e^{2x+7}) \end{aligned}$$

It is possible to start in on this problem with the product rule and then the chain rule, but it's easier if we simplify first. Since $\tan^2x+1=\sec^2 x=\frac{1}{\cos^2x}$, we see $$f(x)=\frac{\cos^2 x}{\cos^2 x}=1$$ for all values of $x$ for which $\cos x$ is nonzero. That is, $f(x)=1$ for every $x$ that is not an integer multiple of $\pi/2$ (and $f(x)$ is not defined when $x$ is an integer multiple of $\pi/2$). Therefore, $f'(x)=0$ for every $x$ on which $f$ exists, and in particular $f'(\pi/4)=0$. Also, $f(\pi/4)=1$, so the tangent line to $f$ at $x=\pi/4$ is the line with slope 0, passing through the point $(\pi/4,1)$: $$y=1$$

The slope of the tangent line is the derivative. If we let $f(x)=\tan x$ and $g(x)=e^{x^2}$, then $f(g(x))=\tan(e^{x^2})$, so $y'=f'(g(x)) \cdot g'(x)$:

$$\begin{aligned} y'&=\sec^2( e^{x^2})\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{ e^{x^2}\} \\ \text{We find ourselves once more in need of the chain rule:} \\ &=\sec^2({e^{x^2}})\cdot{e^{ x^2}}\frac{\mathrm{d}}{\mathrm{d}x}\{ x^2\}\\ &=\sec^2(e^{x^2})\cdot e^{x^2}\cdot 2x \\ \text{Finally, we evaluate this derivative at the point $x=1$:} \\ y'(1)&=\sec^2(e)\cdot e \cdot 2\\ &=2e\sec^2e \end{aligned}$$

Using the Product rule,

$$\begin{aligned} y'&=\frac{\mathrm{d}}{\mathrm{d}x}\{e^{4x}\}\tan x+e^{4x}\sec^2 x \\ \text{and the chain rule:} \\ &=e^{ 4x}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{ 4x\}\cdot\tan x+e^{4x}\sec^2 x\\ &=4e^{4x}\tan x+e^{4x}\sec^2 x \end{aligned}$$