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Categories: Calculus

Table of derivatives

The primary operation in differential calculus is finding a derivative. This table lists derivatives of many functions. In the following, f and g are functions of x, and c is a constant. The set of real numbers is assumed. These formulas are sufficient to differentiate any elementary function.

Contents

1 Rules for differentiation of general functions

2 Derivatives of simple functions

3 Derivatives of exponential and logarithmic functions

4 Derivatives of trigonometric functions

5 Derivatives of hyperbolic functions

Rules for differentiation of general functions

$\left({cf}\right)' = cf'$

$\left({f + g}\right)' = f' + g'$

$\left({f - g}\right)' = f' - g'$

$\left({fg}\right)' = f'g + fg'$

$\left({f \over g}\right)' = {f'g - fg' \over g^2}$

$(f^g)' = f^g\left(f'{g \over f} + g'\ln f\right),\qquad f > 0$

$(f \circ g)' = (f' \circ g)g'$

Derivatives of simple functions

${d \over dx} c = 0$

${d \over dx} x = 1$

${d \over dx} |x| = {x \over |x|} = \sgn x,\qquad x \ne 0$

${d \over dx} x^c = cx^{c-1}$

${d \over dx} \sqrt{x} = {1 \over 2 \sqrt{x}}$

${d \over dx} \left({1 \over x}\right) = -{1 \over x^2}$

Derivatives of exponential and logarithmic functions

${d \over dx} c^x = {c^x \ln c},\qquad c > 0$

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

${d \over dx} \log_c x = {1 \over x \ln c},\qquad c > 0, c \ne 1$

${d \over dx} \ln x = {1 \over x}$

Derivatives of trigonometric functions

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

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

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

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

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

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

${d \over dx} \arcsin x = { 1 \over \sqrt{1 - x^2}}$

${d \over dx} \arccos x = {-1 \over \sqrt{1 - x^2}}$

${d \over dx} \arctan x = { 1 \over 1 + x^2}$

${d \over dx} \arcsec x = { 1 \over |x|\sqrt{x^2 - 1}}$

${d \over dx} \arccot x = {-1 \over 1 + x^2}$

${d \over dx} \arccsc x = {-1 \over |x|\sqrt{x^2 - 1}}$

Derivatives of hyperbolic functions

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

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

${d \over dx} \tanh x = \mbox{sech}^2\,x$

${d \over dx} \,\mbox{sech}\,x = -\tanh x\,\mbox{sech}\,x$

${d \over dx} \,\mbox{coth}\,x = -\,\mbox{csch}^2\,x$

${d \over dx} \,\mbox{csch}\,x = -\,\mbox{coth}\,x\,\mbox{csch}\,x$

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

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

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

${d \over dx} \mbox{sech}^{-1}\,x = { 1 \over x\sqrt{1 - x^2}}$

${d \over dx} \mbox{coth}^{-1}\,x = {-1 \over 1 - x^2}$

${d \over dx} \mbox{csch}^{-1}\,x = {-1 \over |x|\sqrt{1 + x^2}}$

Categories: Calculus

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