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Differentiation Formulas

by Stattler | February 2, 2010

Following is a list of most widely used differentiation formulas. Some other formulas can be found in any standard calculus textbook, such as of James Stewart's CALCULUS Early Transcendentals

General formulas

$\frac{d}{dx} (c) = 0$

$\frac{d}{dx} [c f(x)] = cf'(x)$

$\frac{d}{dx} [f(x) + g(x)] = f'(x) + g'(x)$

$\frac{d}{dx} [f(x) - g(x)]= f'(x) - g'(x)$

$\frac{d}{dx} [f(x)g(x)]= f(x)g'(x) + g(x) f'(x) \quad \text{(aka Product Rule)}$

$\frac{d}{dx} \left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f'(x)-f(x) g'(x)}{[g(x)^2]} \quad \mbox{(aka Quotient Rule)}$

$\frac{d}{dx} f(g(x))=f'(g(x))g'(x) \quad \mbox{(aka Chain Rule)}$

$\frac{d}{dx} x^n=nx^{n-1} \quad \mbox{(aka Power Rule)}$

Exponential and Logarithmic functions

$\frac{d}{dx} e^x= e^x$

$\frac{d}{dx} \ln |x|= \frac{1}{x}$

$\frac{d}{dx} a^x=a^x \ln a$

$\frac{d}{dx} (\log_a x)= \frac{1}{x \ln a}$

Trigonometric functions

$\frac{d}{dx} \sin x= \cos x$

$\frac{d}{dx} \cos x= -\sin x$

$\frac{d}{dx} \tan x= \sec^2 x$

$\frac{d}{dx} \csc x= - \csc x \cot x$

$\frac{d}{dx} \sec x= \sec x \tan x$

$\frac{d}{dx} \cot x= - \csc^2 x$

Inverse Trigonometric functions

$\frac{d}{dx} \sin^{-1}x= \frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx} \cos^{-1}x= -\frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx} \tan^{-1}x= \frac{1}{1+x^2}$

$\frac{d}{dx} \cot^{-1}x= -\frac{1}{1+x^2}$

$\frac{d}{dx} \sec^{-1}x= \frac{1}{x \sqrt{x^2-1}}$

$\frac{d}{dx} \csc^{-1}x= -\frac{1}{x \sqrt{x^2-1}}$

Hyperbolic functions

$\frac{d}{dx}\sinh x= \cosh x$

$\frac{d}{dx}\cosh x= \sinh x$

$\frac{d}{dx}\tanh x= \sec\mbox{h^2} x$

$\frac{d}{dx}\csc\mbox{h} x= -\csc\mbox{h} x \coth x$

$\frac{d}{dx}\sec\mbox{h} x= -\sec\mbox{h} x \tanh x$

$\frac{d}{dx}\coth x= -\csc\mbox{h^2} x$

Inverse Hyperbolic functions

$\frac{d}{dx} \sinh^{-1} x= \frac{1}{\sqrt{1+x^2}}$

$\frac{d}{dx} \cosh^{-1} x= \frac{1}{\sqrt{x^2-1}}$

$\frac{d}{dx} \tanh^{-1} x= \frac{1}{1-x^2}$

$\frac{d}{dx} \cot\mbox{h}^{-1} x= \frac{1}{1-x^2}$

$\frac{d}{dx} \csc\mbox{h}^{-1} x= -\frac{1}{|x| \sqrt{x^2+1}}$

$\frac{d}{dx} \sec\mbox{h}^{-1} x= -\frac{1}{x \sqrt{1-x^2}}$

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