Differentiation rules

Template:Short description This article is a summary of differentiation rules, that is, rules for computing the derivative of a function in calculus.

Elementary rules of differentiation

Unless otherwise stated, all functions are functions of real numbers ($ℝ$) that return real values, although, more generally, the formulas below apply wherever they are well defined, including the case of complex numbers ($ℂ$).

Constant term rule

For any value of $c$, where $c\in ℝ$, if $f(x)$ is the constant function given by $f(x)=c$, then $\frac{df}{dx}=0$.

Proof

Let $c\in ℝ$ and $f(x)=c$. By the definition of the derivative: $$\begin{array}{cc}{f}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{(c)-(c)}{h}\\ & =\underset{h\to 0}{lim}\frac{0}{h}\\ & =\underset{h\to 0}{lim}0\\ & =0.\end{array}$$

This computation shows that the derivative of any constant function is 0.

Intuitive (geometric) explanation

The derivative of the function at a point is the slope of the line tangent to the curve at the point. The slope of the constant function is 0, because the tangent line to the constant function is horizontal and its angle is 0.

In other words, the value of the constant function, $y$, will not change as the value of $x$ increases or decreases.

At each point, the derivative is the slope of a line that is tangent to the curve at that point. Note: the derivative at point A is positive where green and dash–dot, negative where red and dashed, and 0 where black and solid.

Linearity of differentiation

Template:Main For any functions $f$ and $g$ and any real numbers $a$ and $b$, the derivative of the function $h(x)=af(x)+bg(x)$ with respect to $x$ is $h^{\prime }(x)=a{f}^{\prime }(x)+b{g}^{\prime }(x)$.

In Leibniz's notation, this formula is written as: $$\frac{d(af+bg)}{dx}=a\frac{df}{dx}+b\frac{dg}{dx}.$$

Special cases include:

• The constant factor rule:

$$(af{)}^{\prime }=a{f}^{\prime },$$

• The sum rule:

$$(f+g{)}^{\prime }=f^{\prime }+g^{\prime },$$

• The difference rule:

$$(f-g{)}^{\prime }=f^{\prime }-g^{\prime }.$$

Product rule

Template:Main For the functions $f$ and $g$, the derivative of the function $h(x)=f(x)g(x)$ with respect to $x$ is: $$h^{\prime }(x)=(fg{)}^{\prime }(x)=f^{\prime }(x)g(x)+f(x)g^{\prime }(x).$$

In Leibniz's notation, this formula is written: $$\frac{d(fg)}{dx}=g\frac{df}{dx}+f\frac{dg}{dx}.$$

Chain rule

Template:Main The derivative of the function $h(x)=f(g(x))$ is: $$h^{\prime }(x)=f^{\prime }(g(x))\cdot g^{\prime }(x).$$

In Leibniz's notation, this formula is written as: $$\frac{d}{dx}h(x)={\frac{d}{dz}f(z)|}_{z=g(x)}\cdot \frac{d}{dx}g(x),$$ often abridged to: $$\frac{dh(x)}{dx}=\frac{df(g(x))}{dg(x)}\cdot \frac{dg(x)}{dx}.$$

Focusing on the notion of maps, and the differential being a map $\text{D}$, this formula is written in a more concise way as: $$[\text{D}(f\circ g){]}_x=[\text{D}f{]}_{g(x)}\cdot [\text{D}g{]}_x.$$

Inverse function rule

Template:Main If the function $f$ has an inverse function $g$, meaning that $g(f(x))=x$ and $f(g(y))=y$, then: $$g^{\prime }=\frac{1}{f^{\prime }\circ g}.$$

In Leibniz notation, this formula is written as: $$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}.$$

Power laws, polynomials, quotients, and reciprocals

Polynomial or elementary power rule

Template:Main If $f(x)=x^r$, for any real number $r\ne 0$, then: $$f^{\prime }(x)=r{x}^{r-1}.$$

When $r=1$, this formula becomes the special case that, if $f(x)=x$, then $f^{\prime }(x)=1$.

Combining the power rule with the sum and constant multiple rules permits the computation of the derivative of any polynomial.

Reciprocal rule

Template:Main The derivative of $h(x)=\frac{1}{f(x)}$ for any (nonvanishing) function $f$ is: $$h^{\prime }(x)=-\frac{f^{\prime }(x)}{(f(x){)}^2},$$ wherever $f$ is nonzero.

In Leibniz's notation, this formula is written: $$\frac{d\left(\frac{1}{f}\right)}{dx}=-\frac{1}{f^2}\frac{df}{dx}.$$

The reciprocal rule can be derived either from the quotient rule or from the combination of power rule and chain rule.

Quotient rule

Template:Main If $f$ and $g$ are functions, then: $${\left(\frac{f}{g}\right)}^{\prime }=\frac{f^{\prime }g-g^{\prime }f}{g^2},$$ wherever $g$ is nonzero.

This can be derived from the product rule and the reciprocal rule.

Generalized power rule

Template:Main The elementary power rule generalizes considerably. The most general power rule is the functional power rule: for any functions $f$ and $g$, $$(f^g{)}^{\prime }={\left(e^{g\ln f}\right)}^{\prime }=f^g(f^{\prime }\frac{g}{f}+g^{\prime }\ln f),$$ wherever both sides are well defined.

Special cases:

• If $f(x)=x^a$, then $f^{\prime }(x)=a{x}^{a-1}$ when $a$ is any nonzero real number and $x$ is positive.
• The reciprocal rule may be derived as the special case where $g(x)=-1$.

Derivatives of exponential and logarithmic functions

$$\frac{d}{dx}\left(c^{ax}\right)=a{c}^{ax}\ln c,c>0.$$ The equation above is true for all $c$, but the derivative for $c<0$ yields a complex number.

$$\frac{d}{dx}\left(e^{ax}\right)=a{e}^{ax}.$$

$$\frac{d}{dx}\left({\log }_{c}x\right)=\frac{1}{x\ln c},c>1.$$ The equation above is also true for all $c$ but yields a complex number if $c<0$.

$$\frac{d}{dx}(\ln x)=\frac{1}{x},x>0.$$

$$\frac{d}{dx}(\ln |x|)=\frac{1}{x},x\ne 0.$$

$$\frac{d}{dx}(W(x))=\frac{1}{x+e^{W(x)}},x>-\frac{1}{e},$$ where $W(x)$ is the Lambert W function.

$$\frac{d}{dx}\left(x^x\right)=x^x(1+\ln x).$$

$$\frac{d}{dx}(f_1(x{)}^{f_2(x{)}^{{\left(\mathrm{...}\right)}^{f_n(x)}}})=\left[\sum _{k=1}^n\frac{\partial }{\partial x_k}(f_1(x_1{)}^{f_2(x_2{)}^{{\left(\mathrm{...}\right)}^{f_n(x_n)}}})\right]{|}_{x_1=x_2=\mathrm{...}=x_n=x},\text{ if }{f}_{i<n}(x)>0\text{ and }\frac{d{f}_i}{dx}\text{ exists.}$$

Logarithmic derivatives

The logarithmic derivative is another way of stating the rule for differentiating the logarithm of a function (using the chain rule): $$(\ln f{)}^{\prime }=\frac{f^{\prime }}{f},$$ wherever $f$ is positive.

Logarithmic differentiation is a technique which uses logarithms and its differentiation rules to simplify certain expressions before actually applying the derivative.Template:Citation needed

Logarithms can be used to remove exponents, convert products into sums, and convert division into subtraction—each of which may lead to a simplified expression for taking derivatives.

Derivatives of trigonometric functions

$\frac{d}{dx}\sin x=\cos x$	$\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx}\cos x=-\sin x$	$\frac{d}{dx}\arccos x=-\frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx}\tan x={\sec }^{2}x=\frac{1}{{\cos }^{2}x}=1+{\tan }^{2}x$	$\frac{d}{dx}\arctan x=\frac{1}{1+x^2}$
$\frac{d}{dx}\csc x=-\csc x\cot x$	$\frac{d}{dx}\mathrm{arccsc}x=-\frac{1}{|x|\sqrt{x^2-1}}$
$\frac{d}{dx}\sec x=\sec x\tan x$	$\frac{d}{dx}\mathrm{arcsec}x=\frac{1}{|x|\sqrt{x^2-1}}$
$\frac{d}{dx}\cot x=-{\csc }^{2}x=-\frac{1}{{\sin }^{2}x}=-1-{\cot }^{2}x$	$\frac{d}{dx}\mathrm{arccot}x=-\frac{1}{1+x^2}$

The derivatives in the table above are for when the range of the inverse secant is $[0,\pi ]$ and when the range of the inverse cosecant is $[-\frac{\pi }{2},\frac{\pi }{2}]$.

It is common to additionally define an inverse tangent function with two arguments, $\arctan (y,x)$. Its value lies in the range $[-\pi ,\pi ]$ and reflects the quadrant of the point $(x,y)$. For the first and fourth quadrant (i.e., $x>0$), one has $\arctan (y,x>0)=\arctan (\frac{y}{x})$. Its partial derivatives are: $$\frac{\partial \arctan (y,x)}{\partial y}=\frac{x}{x^2+y^2}\text{and}\frac{\partial \arctan (y,x)}{\partial x}=\frac{-y}{x^2+y^2}.$$

Derivatives of hyperbolic functions

$\frac{d}{dx}\sinh x=\cosh x$	$\frac{d}{dx}\mathrm{arsinh}x=\frac{1}{\sqrt{1+x^2}}$
$\frac{d}{dx}\cosh x=\sinh x$	$\frac{d}{dx}\mathrm{arcosh}x=\frac{1}{\sqrt{x^2-1}}$
$\frac{d}{dx}\tanh x={\mathrm{sech}}^{2}x=1-{\tanh }^{2}x$	$\frac{d}{dx}\mathrm{artanh}x=\frac{1}{1-x^2}$
$\frac{d}{dx}\mathrm{csch}x=-\mathrm{csch}x\coth x$	$\frac{d}{dx}\mathrm{arcsch}x=-\frac{1}{|x|\sqrt{1+x^2}}$
$\frac{d}{dx}\mathrm{sech}x=-\mathrm{sech}x\tanh x$	$\frac{d}{dx}\mathrm{arsech}x=-\frac{1}{x\sqrt{1-x^2}}$
$\frac{d}{dx}\coth x=-{\mathrm{csch}}^{2}x=1-{\coth }^{2}x$	$\frac{d}{dx}\mathrm{arcoth}x=\frac{1}{1-x^2}$

Derivatives of special functions

Gamma function

Template:Main $$\mathrm{\Gamma }(x)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt$$ $$\begin{array}{cc}{\mathrm{\Gamma }}^{\prime }(x)& ={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}\ln tdt\\ & =\mathrm{\Gamma }(x)(\sum _{n=1}^{\mathrm{\infty }}(\ln (1+\frac{1}{n})-\frac{1}{x+n})-\frac{1}{x})\\ & =\mathrm{\Gamma }(x)\psi (x),\end{array}$$ with $\psi (x)$ being the digamma function, expressed by the parenthesized expression to the right of $\mathrm{\Gamma }(x)$ in the line above.

Riemann zeta function

Template:Main $$\zeta (x)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^x}$$ $$\begin{array}{cc}{\zeta }^{\prime }(x)& =-\sum _{n=1}^{\mathrm{\infty }}\frac{\ln n}{n^x}=-\frac{\ln 2}{2^x}-\frac{\ln 3}{3^x}-\frac{\ln 4}{4^x}-\cdots \\ & =-\sum _{p\text{ prime}}\frac{p^{-x}\ln p}{(1-p^{-x}{)}^2}\prod _{q\text{ prime},q\ne p}\frac{1}{1-q^{-x}}\end{array}$$

Derivatives of integrals

Template:Main Suppose that it is required to differentiate with respect to $x$ the function: $$F(x)={\int }_{a(x)}^{b(x)}f(x,t)dt,$$

where the functions $f(x,t)$ and $\frac{\partial }{\partial x}f(x,t)$ are both continuous in both $t$ and $x$ in some region of the $(t,x)$ plane, including $a(x)\le t\le b(x)$, where $x_0\le x\le x_1$, and the functions $a(x)$ and $b(x)$ are both continuous and both have continuous derivatives for $x_0\le x\le x_1$. Then, for $x_0\le x\le x_1$: $$F^{\prime }(x)=f(x,b(x))b^{\prime }(x)-f(x,a(x))a^{\prime }(x)+{\int }_{a(x)}^{b(x)}\frac{\partial }{\partial x}f(x,t)dt.$$

This formula is the general form of the Leibniz integral rule and can be derived using the fundamental theorem of calculus.

Derivatives to nth order

Some rules exist for computing the $n$th derivative of functions, where $n$ is a positive integer, including:

Faà di Bruno's formula

Template:Main If $f$ and $g$ are $n$-times differentiable, then: $$\frac{d^n}{d{x}^n}[f(g(x))]=n!\sum _{\{k_m\}}{f}^{(r)}(g(x))\prod _{m=1}^n\frac{1}{k_m!}{(g^{(m)}(x))}^{k_m},$$ where $r={\sum }_{m=1}^{n-1}{k}_m$ and the set $\{k_m\}$ consists of all non-negative integer solutions of the Diophantine equation ${\sum }_{m=1}^{n}m{k}_m=n$.

General Leibniz rule

Template:Main If $f$ and $g$ are $n$-times differentiable, then: $$\frac{d^n}{d{x}^n}[f(x)g(x)]=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})\frac{d^{n-k}}{d{x}^{n-k}}f(x)\frac{d^k}{d{x}^k}g(x).$$

See also

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Sources and further reading

These rules are given in many books, both on elementary and advanced calculus, in pure and applied mathematics. Those in this article (in addition to the above references) can be found in:

• Mathematical Handbook of Formulas and Tables (3rd edition), S. Lipschutz, M.R. Spiegel, J. Liu, Schaum's Outline Series, 2009, Template:ISBN.
• The Cambridge Handbook of Physics Formulas, G. Woan, Cambridge University Press, 2010, Template:ISBN.
• Mathematical methods for physics and engineering, K.F. Riley, M.P. Hobson, S.J. Bence, Cambridge University Press, 2010, Template:ISBN
• NIST Handbook of Mathematical Functions, F. W. J. Olver, D. W. Lozier, R. F. Boisvert, C. W. Clark, Cambridge University Press, 2010, Template:ISBN.

External links

• Derivative calculator with formula simplification
• The table of derivatives with animated proves