합성함수의 미분법

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 $x^2 + y^2 = \sin^2t + \cos^2t = 1$.

Let $f(x)=e^{x+x^2}$ and $g(x)=1+x$. Then $f(0)=g(0)=1$.

$f'(x)=(1+2x)e^{x+x^2}$ and $g'(x)=1$. When $x>0$, $$f'(x)=(1+2x)e^{x+x^2}>1\cdot e^{x+x^2}=e^{x+x^2}>e^{0+0^2}=1=g'(x).$$ Since $f(0)=g(0)$, and $f'(x)>g'(x)$ for all $x>0$, that means $f$ and $g$ start at the same place, but $f$ always grows faster. Therefore, $f(x)>g(x)$ for all $x>0$.

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}$$