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# 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 with respect to x. The set of real numbers is assumed. These formulas are sufficient to differentiate any elementary function.

### Rules for differentiation of general functions

${d \over dx} cf(x) = c{d \over dx} f(x)$

${d \over dx} (f(x) + g(x)) = {d \over dx} f(x) + {d \over dx} g(x)$

${d \over dx} f(x)g(x) = {d \over dx}f(x) \cdot g(x) + f(x) \cdot {d \over dx}g(x)$

${d \over dx} {f(x) \over g(x)} = {{d \over dx} f(x) \cdot g(x) - f(x) \cdot {d \over dx} g(x) \over (g(x))^2}$

${d \over dx} f(x)^{g(x)} = f(x)^{g(x)}\left({d \over dx}f(x) \cdot {g(x) \over f(x)} + {d \over dx}g(x) \cdot \ln f(x)\right),\qquad f(x) > 0$

${d \over dx} f(g(x)) = {d \over dg} f(g(x)) {d \over dx} g(x)$

### Derivatives of Simple and Polynomial Functions

${d \over dx} c = 0$

${d \over dx} x = 1$

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

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

### Derivatives of Exponential and Logarithmic Functions

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

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

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

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

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

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

### Derivatives of Trigonometric Functions

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

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

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

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

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

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

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

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

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

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

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

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

### 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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