합성함수의 미분법

$$\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}$$