Methods of Differentiation

$f'(x)=\lim_{h\to0}\frac{(x+h)^n-x^n}{h}$
$(x+h)^n=x^n+nhx^{n-1}+\frac{n(n-1)}{2}h^2x^{n-2}+\cdots$
Therefore,
$f'(x)=\lim_{h\to0}\frac{x^n+nhx^{n-1}+\frac{n(n-1)}{2}h^2x^{n-2}+\cdots-x^n}{h}$
$f'(x)=\lim_{h\to0}\frac{nhx^{n-1}+\frac{n(n-1)}{2}h^2x^{n-2}+\cdots}{h}$
$f'(x)=\lim_{h\to0}\left(nx^{n-1}+\frac{n(n-1)}{2}hx^{n-2}+\cdots\right)$
As $h\to0$, all terms containing $h$ vanish.
Hence,
$f'(x)=nx^{n-1}$

$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$
$f'(x)=\lim_{h\to0}\frac{a^{x+h}-a^x}{h}$
$a^{x+h}=a^x a^h$
Therefore,
$f'(x)=\lim_{h\to0}\frac{a^x(a^h-1)}{h}$
$f'(x)=a^x\lim_{h\to0}\frac{a^h-1}{h}$
Using the standard limit,
$\lim_{h\to0}\frac{a^h-1}{h}=\ln a$
Hence,
$f'(x)=a^x\ln a$

$f'(x)=\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}$
Using the identity,
$\sin(x+h)-\sin x=2\cos\left(x+\frac{h}{2}\right)\sin\frac{h}{2}$
Therefore,
$f'(x)=\lim_{h\to0}\frac{2\cos\left(x+\frac{h}{2}\right)\sin\frac{h}{2}}{h}$
Rewrite $h$ as $2\cdot\frac{h}{2}$:
$f'(x)=\lim_{h\to0}\cos\left(x+\frac{h}{2}\right)\frac{\sin\frac{h}{2}}{\frac{h}{2}}$
As $h\to0$, $\cos\left(x+\frac{h}{2}\right)\to\cos x$ and $\frac{\sin\frac{h}{2}}{\frac{h}{2}}\to1$
Hence,
$f'(x)=\cos x$

$D(e^\pi)=0$ because $e^\pi$ is constant.
$D(x^e)=e x^{e-1}$ by the power rule.

Rewrite the function as $\csc x$.
Then apply the standard derivative of $\csc x$.
$\frac{d}{dx}\left(\frac{1}{\sin x}\right)=\frac{d}{dx}(\csc x)=-\csc x\cot x$.

$\frac{\sin 2x}{1+\cos 2x}=\frac{2\sin x\cos x}{2\cos^2x}$.
This simplifies to $\tan x$.
Therefore, $D\left(\frac{\sin 2x}{1+\cos 2x}\right)=D(\tan x)=\sec^2x$.

Since $\tan(\tan^{-1}x)=x$, the given function is simply $x$.
Therefore, $D(\tan(\tan^{-1}x))=D(x)=1$.

$y=\left(x^{a-b}\right)^{a+b}\left(x^{b-c}\right)^{b+c}\left(x^{c-a}\right)^{c+a}$.
$y=x^{a^2-b^2}x^{b^2-c^2}x^{c^2-a^2}$.
$y=x^{a^2-b^2+b^2-c^2+c^2-a^2}$.
$y=x^0=1$.
Therefore, $\frac{dy}{dx}=0$.

Given $y=e^x+3\ln x-4\tan^{-1}x+\sin 3x+4\sin^3x$.
Use $\sin 3x=3\sin x-4\sin^3x$.
Then $y=e^x+3\ln x-4\tan^{-1}x+3\sin x$.
Differentiate term by term.
$\frac{dy}{dx}=e^x+\frac{3}{x}-\frac{4}{1+x^2}+3\cos x$.

$y=e^x+3\ln x-4\sin x$.
Differentiate each term separately.
$\frac{dy}{dx}=e^x+\frac{3}{x}-4\cos x$.

Let $u=x$ and $v=\sin^{-1}x$.
Then $u'=1$ and $v'=\frac{1}{\sqrt{1-x^2}}$.
By product rule, $\frac{dy}{dx}=1\cdot\sin^{-1}x+x\cdot\frac{1}{\sqrt{1-x^2}}$.
Thus, $\frac{dy}{dx}=\sin^{-1}x+\frac{x}{\sqrt{1-x^2}}$.

Let $f=x^2$, $g=e^x$, $h=\ln x$.
Then $f'=2x$, $g'=e^x$, $h'=\frac{1}{x}$.
$\frac{dy}{dx}=2x e^x\ln x+x^2e^x\ln x+x^2e^x\frac{1}{x}$.
$\frac{dy}{dx}=2x e^x\ln x+x^2e^x\ln x+x e^x$.
Factorising gives $x e^x\left((x+2)\ln x+1\right)$.

Let $u=1+x$, $v=(3+x^2)^{1/2}$, $w=(9+x^3)^{1/3}$.
Then $f'=u'vw+uv'w+uvw'$.
At $x=-1$, $u=0$, so the second and third terms vanish.
Also, $u'=1$.
$v(-1)=(3+1)^{1/2}=2$.
$w(-1)=(9-1)^{1/3}=8^{1/3}=2$.
Thus, $f'(-1)=1\cdot2\cdot2=4$.

Let $N=x^3+x^2+1$ and $D=x^2+1$
Then $N'=3x^2+2x$ and $D'=2x$
By quotient rule,
$\frac{dy}{dx}=\frac{(x^2+1)(3x^2+2x)-(x^3+x^2+1)(2x)}{(x^2+1)^2}$
Expand numerator:
$(x^2+1)(3x^2+2x)=3x^4+2x^3+3x^2+2x$
$(x^3+x^2+1)(2x)=2x^4+2x^3+2x$
Subtract:
$3x^4+2x^3+3x^2+2x-(2x^4+2x^3+2x)=x^4+3x^2$
Therefore,
$\frac{dy}{dx}=\frac{x^4+3x^2}{(x^2+1)^2}=\frac{x^2(x^2+3)}{(x^2+1)^2}$

Given $y=\frac{\sin^{-1}x-\cos^{-1}x}{\sin^{-1}x+\cos^{-1}x}$
Use $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$
Also $\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x$
Numerator becomes $\sin^{-1}x-\left(\frac{\pi}{2}-\sin^{-1}x\right)=2\sin^{-1}x-\frac{\pi}{2}$
Thus $y=\frac{2\sin^{-1}x-\frac{\pi}{2}}{\frac{\pi}{2}}$
$y=\frac{4}{\pi}\sin^{-1}x-1$
Differentiate:
$\frac{dy}{dx}=\frac{4}{\pi}\cdot\frac{1}{\sqrt{1-x^2}}$
$\frac{dy}{dx}=\frac{4}{\pi\sqrt{1-x^2}}$

$f(0)=0$
$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}$
$f'(0)=\lim_{h\to0}\frac{h^{1/3}\sin h}{h}$
$f'(0)=\lim_{h\to0}h^{1/3}\cdot\frac{\sin h}{h}$
Since $\frac{\sin h}{h}\to1$ and $h^{1/3}\to0$,
$f'(0)=0$

Let $u=\log x$
Then $y=\tan u$
$\frac{dy}{du}=\sec^2u$
$\frac{du}{dx}=\frac{1}{x}$
By chain rule,
$\frac{dy}{dx}=\sec^2(\log x)\cdot\frac{1}{x}$
$\frac{dy}{dx}=\frac{\sec^2(\log x)}{x}$

Let $u=x^2$
Then $\frac{du}{dx}=2x$
$\frac{d}{dx}\sin(x^2)=\cos(x^2)\cdot2x$
$\frac{d}{dx}\sin(x^2)=2x\cos(x^2)$

Let $u=\sin(x^2)$
Then $\frac{du}{dx}=\cos(x^2)\cdot2x$
$\frac{d}{dx}\log(\sin(x^2))=\frac{2x\cos(x^2)}{\sin(x^2)}$
$\frac{d}{dx}\log(\sin(x^2))=2x\cot(x^2)$

Let $u=\log(\sin(x^2))$
Then $y=\tan^{-1}u$
$\frac{dy}{du}=\frac{1}{1+u^2}$
Now find $\frac{du}{dx}$
Let $v=\sin(x^2)$
Then $u=\log v$
$\frac{du}{dx}=\frac{1}{v}\cdot\frac{dv}{dx}$
$\frac{dv}{dx}=\cos(x^2)\cdot2x$
Thus $\frac{du}{dx}=\frac{2x\cos(x^2)}{\sin(x^2)}=2x\cot(x^2)$
Therefore,
$\frac{dy}{dx}=\frac{2x\cot(x^2)}{1+(\log(\sin(x^2)))^2}$