List of derivatives

We have

$$f'(x)=\lim_{y\to x}\frac{f(y)-f(x)}{y-x}$$

For many elementary functions this type of limit is very easy to be found. We need some algebra but sometimes you have to use some remarkable limits.

$$f(x)=c\:f'(x)=0\; as \:\lim_{y\to x}\frac{c-c}{y-x}=0$$
$$f(x)=x\: f'(x)=1\:as \lim_{y\to x}\frac{y-x}{y-x}=1$$
$$(x^n)'=nx^{n-1}$$

as

$$\lim_{y\to x}\frac{y^n-x^n}{y-x}=\lim_{y\to x}y^{n-1}+y^{n-2}x+...+yx^{n-2}+x^{n-1}=nx^{n-1}$$
$$(lnx)'=\frac{1}{x}$$
$$(\sqrt x)'=\frac{1}{2\sqrt x};\lim_{y\to x}\frac{\sqrt y-\sqrt x}{y-x}=\lim_{y\to x}\frac{1}{\sqrt y+\sqrt x}=\frac{1}{2\sqrt x}$$
$$(\sqrt[3] x)'=\frac{1}{3\sqrt[3] {x^2}};\lim_{y\to x}\frac{\sqrt[3] y-\sqrt[3] x}{y-x}=\lim_{y\to x}\frac{1}{\sqrt [3]{y^2}+\sqrt[3 ]{yx}+\sqrt [3]{x^2}}=\frac{1}{3\sqrt[3]{ x^2}}$$
$$(\sqrt[n] x)'=\frac{1}{n\sqrt[n] {x^{n-1}}}$$
$$(e^x)'=e^x$$
$$(a^x)'=(e^{x\ln a})'=\ln ae^{x\ln a}=a^x ln a$$
$$(e^{f(x)})'=e^{f(x)}f'(x)$$
$$ (\sin x)'=\cos x; (\sin ax)'=a\cos ax$$

Proof

$$ (\sin x)'=\lim_{y\to x}\frac{\sin y-\sin x}{y-x}=\lim_{y\to x}\frac{2\sin\frac{y-x}{2}\cos \frac{y+x}{2}}{y-x}=\lim_{y\to x}\frac{\sin\frac{y-x}{2}}{\frac{y-x}{2}}\cos\frac{y+x}{2}=\cos x$$
$$ (\cos x)'=-\sin x;(\cos ax)'=-a\sin ax$$

Proof

$$ (\cos x)'=\lim_{y\to x}\frac{\cos y-\cos x}{y-x}=\lim_{y\to x}-\frac{2\sin\frac{y-x}{2}\sin  \frac{y+x}{2}}{y-x}=-\lim_{y\to x}\frac{\sin\frac{y-x}{2}}{\frac{y-x}{2}}\sin\frac{y+x}{2}=-\sin x$$
$$(\tan x)'=1+\tan^2x=\frac{1}{\cos^2x};(\tan ax)'=a(1+\tan^2ax)=\frac{a}{\cos^2ax}$$
$$ (\cot x)'=-(1+\cot^2x)=-\frac{1}{\sin^2x};(\cot ax)'=-a(1+\cot^2ax)=-\frac{a}{\sin^2ax}$$
$$(\arcsin x)'=\frac{1}{\sqrt{1-x^2}};(\arcsin ax)'=\frac{a}{\sqrt{1-a^2x^2}}$$
$$(\arccos x)'=-\frac{1}{\sqrt{1-x^2}};(\arccos ax)'=-\frac{a}{\sqrt{1-a^2x^2}}$$
$$(\arctan x)'=\frac{1}{1+x^2};(\arctan ax)'=\frac{a}{1+a^2x^2}$$
$$(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)$$
$$(\frac{f(x)}{g(x)})'=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}$$
$$(g(f(x)))'=g'(f(x)f'(x)$$
$$ (\sin \ln x)'=\frac{1}{x}\cos\ln x$$
$$(\cos\ln x)'=-\frac{1}{x}\sin \ln x$$
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