합성함수의 미분법

We start with the chain rule:

$$\begin{aligned} y'&=-\sin\big( x^2+\sqrt{x^2+1}\big)\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left\{ x^2+\sqrt{x^2+1}\right\}\\ &=-\sin\big(x^2+\sqrt{x^2+1}\big)\cdot\left(2x+\frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{x^2+1}\right\}\right) \\ \text{and find ourselves in need of chain rule a second time:} \\ &=-\sin\big(x^2+\sqrt{x^2+1}\big)\cdot\left(2x+\dfrac{1}{2\sqrt{ x^2+1}}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left\{ x^2+1\right\}\right)\\ &=-\sin\big(x^2+\sqrt{x^2+1}\big)\cdot\left(2x+\dfrac{2x}{2\sqrt{x^2+1}}\right) \end{aligned}$$

$$\begin{aligned} y&=(1+x^2)\cos^2 x \\ \text{Using the product rule,} \\ y'&=(2x)\cos^2 x + (1+x^2)\frac{\mathrm{d}}{\mathrm{d}x}\{\cos^2 x\} \\ \text{Here, we'll need to use the chain rule. Remember $\cos^2 x = [\cos x]^2$.} \\ &=2x\cos^2x+(1+x^2) 2 \cos x \cdot \frac{\mathrm{d}}{\mathrm{d}x}\{ \cos x\}\\ &=2x\cos^2x+(1+x^2) 2\cos x \cdot (-\sin x)\\ &=2x\cos^2x-2(1+x^2) \sin x\cos x \end{aligned}$$

We use the quotient rule, noting by the chain rule that $\frac{\mathrm{d}}{\mathrm{d}x}\{e^{3x}\}=3e^{3x}$:

$$\begin{aligned} y'&=\frac{(1+x^2)\cdot3e^{3x}-e^{3x}(2x)}{(1+x^2)^2}\\ &=\frac{e^{3x}(3x^2-2x+3)}{(1+x^2)^2} \end{aligned}$$

By the chain rule,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{h\left(x^2\right)\right\}&=h'(x^2)\cdot 2x \end{aligned}$$

Using the product rules and the result above,

$$\begin{aligned} g'(x)&=3x^2h(x^2)+x^3h'(x^2)2x \\ \text{Plugging in $x=2$:} \\ g'(2)&=3(2^2)h(2^2)+2^3h'(2^2)2\times 2\\ &=12h(4)+32h'(4)=12\times 2-32\times 2\\ &=-40 \end{aligned}$$

Let $f(x)=xe^{-(x^2-1)/2}=xe^{(1-x^2)/2}$. Then, using the product rule,

$$\begin{aligned} f'(x)&=e^{(1-x^2)/2}+x\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{(1-x^2)/2}\right\} \\ \text{Here, we need the chain rule:} \\ &=e^{(1-x^2)/2}+x\cdot e^{ (1-x^2)/2}\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \frac{1}{2}(1-x^2)\right\}\\ &=e^{(1-x^2)/2}+x\cdot e^{{(1-x^2)/2}}\cdot (-x)\\ &=(1-x^2)e^{(1-x^2)/2} \end{aligned}$$

There is no power of $e$ that is equal to zero; so if the product above is zero, it must be that $1-x^2=0$. This happens for $x=\pm 1$. On the curve, when $x=1$, $y=1$, and when $x=-1$, $y=-1$. So the points are $(1,1)$ and $(-1,-1)$.

The question asks when $s'(t)$ is negative. So, we start by differentiating. Using the chain rule:

$$\begin{aligned} s'(t)&=\cos\left( \frac{1}{t}\right)\cdot\frac{\mathrm{d}}{\mathrm{d}t}\left\{ \frac{1}{t}\right\}\\ &=\cos\left({\frac{1}{t}}\right)\cdot\frac{-1}{t^2} \end{aligned}$$

When $t\ge1$, $\frac{1}{t}$ is between 0 and 1. Since $\cos \theta$ is positive for $0 \leq \theta < \pi/2$, and $\pi/2 >1$, we see that $\cos\left(\frac{1}{t}\right)$ is positive for the entire domain of $s(t)$. Also, $\frac{-1}{t^2}$ is negative for the entire domain of the function. We conclude that $s'(t)$ is negative for the entire domain of $s(t)$, so the particle is always moving in the negative direction.

We present two solutions: one where we dive right in and use the quotient rule, and another where we simplify first and use the product rule.

• Solution 1: We begin with the quotient rule:

$$\begin{aligned} f'(x) &= \frac{\cos^3(5x-7)\frac{\mathrm{d}}{\mathrm{d}x}\{e^x\}-e^x\frac{\mathrm{d}}{\mathrm{d}x}\{\cos^3(5x-7)\}}{\cos^6(5x-7)}\\ &= \frac{\cos^3(5x-7)e^x-e^x\frac{\mathrm{d}}{\mathrm{d}x}\{\cos^3(5x-7)\}}{\cos^6(5x-7)} \\ \text{Now, we use the chain rule. Since $\cos^3(5x-7)=[\cos(5x-7)]^3$, our ``outside" function is $g(x)=x^3$, and our ``inside" function is $h(x)=\cos(5x-1)$.} \\ &= \frac{\cos^3(5x-7)e^x-e^x\cdot3 \cos^2 (5x-7) \cdot \frac{\mathrm{d}}{\mathrm{d}x}\{ \cos(5x-7)\}}{\cos^6(5x-7)} \\ \text{We need the chain rule again!} \\ &= \frac{\cos^3(5x-7)e^x-e^x\cdot3{\cos}^2{(5x-7)} \cdot[{-\sin( 5x-7)\cdot \frac{\mathrm{d}}{\mathrm{d}x}\{ 5x-7\}}]}{\cos^6(5x-7)}\\ &= \frac{\cos^3(5x-7)e^x-e^x\cdot3{\cos}^2{(5x-7)} \cdot[{-\sin({5x-7})\cdot5}]}{\cos^6(5x-7)} \\ \text{We finish by simplifying:} \\ &= \frac{e^x\cos^2(5x-7)\left(\cos(5x-7)+15\sin(5x-7)\right)}{\cos^6(5x-7)}\\ &=e^x \frac{\cos(5x-7)+15\sin(5x-7)}{\cos^4(5x-7)}\\ &=e^x(\sec^3(5x-7)+15\tan(5x-7)\sec^3(5x-7))\\ &=e^x\sec^3(5x-7)(1+15\tan(5x-7)) \end{aligned}$$

• Solution 2: We simplify to avoid the quotient rule:

$$\begin{aligned} f(x)&=\dfrac{e^{x}}{\cos^3 (5x-7)}\\ &=e^x\sec^3(5x-7) \\ \text{Now we use the product rule to differentiate:} \\ f'(x)&=e^x\sec^3(5x-7)+e^x\frac{\mathrm{d}}{\mathrm{d}x}\{\sec^3(5x-7)\} \\ \text{Here, we'll need the chain rule. Since $\sec^3(5x-7)=[\sec (5x-7)]^3$, our ``outside" function is $g(x)=x^3$ and our ``inside" function is $h(x)=\sec(5x-7)$, so that $g(h(x))=[\sec(5x-7)]^3=\sec^3(5x-7)$.} \\ &=e^x\sec^3(5x-7)+e^x\cdot3~ \sec^2 (5x-7) \cdot \frac{\mathrm{d}}{\mathrm{d}x}\{ \sec(5x-7)\} \\ \text{We need the chain rule again! Recall $\frac{\mathrm{d}}{\mathrm{d}x}\{\sec x\}=\sec x \tan x$.} \\ &=e^x\sec^3(5x-7)+e^x\cdot3~{\sec}^2{(5x-7)} \cdot {\sec( 5x-7)\tan( 5x-7)\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{ 5x-7\}}\\ &=e^x\sec^3(5x-7)+e^x\cdot3~{\sec}^2{(5x-7)} \cdot {\sec({5x-7})\tan({5x-7})\cdot 5} \\ \text{We finish by simplifying:} \\ &=e^x\sec^3(5x-7)(1+15\tan({5x-7})) \end{aligned}$$

At time $t$, the particle is at the point $\big(x(t),y(t)\big)$, with $x(t)=\cos t$ and $y(t)=\sin t$. Over time, the particle traces out a curve; let's call that curve $y=f(x)$. Then $y(t) = f\big(x(t)\big)$, so the slope of the curve at the point $\big(x(t),y(t)\big)$ is $f'\big(x(t)\big)$. You are to determine the values of $t$ for which $f'\big(x(t)\big)=-1$.

By the chain rule

$$\begin{aligned} y'(t) = f'\big(x(t)\big) \cdot x'(t) \end{aligned}$$

Substituting in $x(t)=\cos t$ and $y(t)=\sin t$ gives

$$\begin{aligned} \cos t = f'\big(x(t)\big) \cdot \big(-\sin t\big) \end{aligned}$$

so that

$$\begin{aligned} f'\big(x(t)\big) = -\frac{\cos t}{\sin t} \end{aligned}$$

is $-1$ precisely when $\sin t = \cos t$. This happens whenever $t = \frac{\pi}{4}$.

Remark: the path traced by the particle is a semicircle. You can think about the point on the unit circle with angle t, or you can notice that $x^2 + y^2 = \sin^2t + \cos^2t = 1$.

Since $\sin 2x$ and $2\sin x \cos x$ are the same function, they have the same derivative.

$$\begin{aligned} \sin 2x &= 2\sin x \cos x\\ \Rightarrow \frac{\mathrm{d}}{\mathrm{d}x}\{\sin 2x\}&=\frac{\mathrm{d}}{\mathrm{d}x}\{2\sin x \cos x\}\\ 2\cos 2x &=2[\cos^2x-\sin^2x]\\ \cos 2x &=\cos^2x-\sin^2 x \end{aligned}$$

We conclude $\cos 2x =\cos^2x-\sin^2 x$, which is another common trig identity.

Remark: if we differentiate both sides of this equation, we get the original identity back.

$$\begin{aligned} f(x)&=\sqrt[3]{\dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }}\\ &=\left({\dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }}\right)^{\frac{1}{3}} \\ \text{To begin the differentiation, we can choose our ``outside" function to be $g(x)=x^{\frac{1}{3}}$, and our ``inside" function to be $h(x)=\dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }$. Then $f(x)=g(h(x))$, so $f'(x)=g'(h(x))\cdot h'(x)=\frac{1}{3}(h(x))^{-\frac{2}{3}}h'(x)$:} \\ f'(x)&=\frac{1}{3}\left( \dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }\right)^{\frac{-2}{3}}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }\right\}\\ &=\frac{1}{3} \left( \dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}} \right)^{\frac{2}{3}} \cdot \frac{\mathrm{d}}{\mathrm{d}x}\left\{ \dfrac{e^{\csc x^2}}{ \sqrt{x^3-9} \tan x }\right\} \\ \text{This leads us to use the quotient rule:} \\ &=\frac{1}{3}\left( \dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}} \right)^{\frac{2}{3}} \left(\frac{ \sqrt{x^3-9}\tan x \frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{\csc x^2}\right\}-e^{\csc x^2}\frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{x^3-9}\tan x \right\}}{(\tan^2 x)(x^3-9) }\right) \\ \text{Let's figure out those two derivatives on their own, then plug them in. Using the chain rule twice:} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{\csc x^2}\right\}&=e^{ \csc x^2}\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \csc x^2\right\}=e^{\csc x^2}\cdot (-\csc( x^2)\cot( x^2))\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{ x^2\}\\&=-2xe^{\csc x^2}\csc(x^2)\cot(x^2) \\ \text{For the other derivative, we start with the product rule, then chain:} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{ \sqrt{x^3-9} \tan x \right\}&= \frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{x^3-9}\right\}\cdot\tan x+\sqrt{x^3-9}\sec^2 x\\ &=\frac{1}{2\sqrt{ x^3-9}}\frac{\mathrm{d}}{\mathrm{d}x}\left\{ x^3-9\right\}\cdot \tan x+\sqrt{x^3-9}\sec^2 x\\ &=\frac{3x^2 \tan x}{2\sqrt{{x^3-9}}}+\sqrt{x^3-9}\sec^2 x \\ \text{Now, we plug these into our equation for $f'(x)$:} \\ f'(x)&=\frac{1}{3}\left( \dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}} \right)^{\frac{2}{3}} \left(\frac{ \sqrt{x^3-9}\tan x \frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{\csc x^2}\right\}-e^{\csc x^2} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{x^3-9}\tan x \right\}}{(\tan^2 x)(x^3-9) }\right)\\ &=\frac{1}{3}\left( \dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}} \right)^{\frac{2}{3}}\cdot\\ &~\left(\frac{ \sqrt{x^3-9}\tan x (-2x)e^{\csc x^2}\csc(x^2)\cot(x^2)-e^{\csc x^2} \left(\frac{3x^2\tan x}{2\sqrt{{x^3-9}}}+\sqrt{x^3-9}\sec^2 x\right)}{(\tan^2 x)(x^3-9) }\right) \end{aligned}$$