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.

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}
{d \over dx} \left\over (\ln x)^n} \right = \frac{-n}{x\,(\ln x)^{n+1}}

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} \sin^{-1} x = { 1 \over \sqrt{1 - x^2}}
{d \over dx} \cos^{-1} x = {-1 \over \sqrt{1 - x^2}}
{d \over dx} \tan^{-1} x = { 1 \over 1 + x^2}
{d \over dx} \sec^{-1} x = { 1 \over |x|\sqrt{x^2 - 1}}
{d \over dx} \cot^{-1} x = {-1 \over 1 + x^2}
{d \over dx} \csc^{-1} 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}}

 

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