미분법 전체 문제

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 $(\pi/4,1)$: $$y=1$$

Velocity is the derivative of position with respect to time. So, the velocity of the particle is given by $s'(t)$. We need to find $s'(t)$, and determine when it is zero.

To differentiate, we us the chain rule.

$$\begin{aligned} s'(t)&=e^{ t^3-7t^2+8t}\cdot\frac{\mathrm{d}}{\mathrm{d}t}\{ t^3-7t^2+8t\}\\ &=e^{{t^3-7t^2+8t}}\cdot(3t^2-14t+8) \\ \text{To determine where this function is zero, we factor:} \\ &=e^{{t^3-7t^2+8t}}\cdot(3t-2)(t-4) \end{aligned}$$

So, the velocity is zero when $e^{{t^3-7t^2+8t}}=0$, when $3t-2=0$, and when $t-4=0$. Since $e^{{t^3-7t^2+8t}}$ is never zero, this tells us that the velocity is zero precisely when $t=\frac{2}{3}$ or $t=4$.

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,

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

Using the quotient rule,

$$\begin{aligned} f'(x)&=\frac{(3x^2)(1+e^{3x})-(x^3)\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{1+e^{3x}\}}{{(1+e^{3x})}^2} \\ \text{Now, the chain rule:} \\ &=\frac{(3x^2)(1+e^{3x})-(x^3)(3e^{3x})}{{(1+e^{3x})}^2} \\ \text{So, when $x=1$:} \\ f'(1)&=\frac{3(1+e^{3})-3e^{3}}{{(1+e^{3})}^2}=\frac{3}{{(1+e^{3})}^2} \end{aligned}$$

This requires us to apply the chain rule twice.

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

This requires us to apply the chain rule twice.

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x} \left\{\sin(e^{5x}) \right\} &= {\cos\left( e^{5x}\right)} \cdot \frac{\mathrm{d}}{\mathrm{d}x} \left\{ e^{5x}\right\}\\ &= \cos(e^{5x}) (e^{ 5x}) \cdot \frac{\mathrm{d}}{\mathrm{d}x}\{ 5x\} \\ &= \cos(e^{5x}) (e^{5x}) \cdot5 \end{aligned}$$

We'll use the chain rule twice.

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

We 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.