# Chain rule  Main Article Discussion Related Articles  [?] Bibliography  [?] External Links  [?] Citable Version  [?] This editable Main Article is under development and subject to a disclaimer. [edit intro]

In calculus, the chain rule describes the derivative of a "function of a function": the composition of two function, where the output z is a given function of an intermediate variable y which is in turn a given function of the input variable x.

Suppose that y is given as a function $\,y=g(x)$ and that z is given as a function $\,z=f(y)$ . The rate at which z varies in terms of y is given by the derivative $\,f'(y)$ , and the rate at which y varies in terms of x is given by the derivative $\,g'(x)$ . So the rate at which z varies in terms of x is the product $\,f'(y)\cdot g'(x)$ , and substituting $\,y=g(x)$ we have the chain rule

$(f\circ g)'=(f'\circ g)\cdot g'.\,$ In order to convert this to the traditional (Leibniz) notation, we notice

$z(y(x))\quad \Longleftrightarrow \quad z\circ y(x)$ and

$(z\circ y)'=(z'\circ y)\cdot y'\quad \Longleftrightarrow \quad {\frac {\mathrm {d} z(y(x))}{\mathrm {d} x}}={\frac {\mathrm {d} z(y)}{\mathrm {d} y}}\,{\frac {\mathrm {d} y(x)}{\mathrm {d} x}}.\,$ .

In mnemonic form the latter expression is

${\frac {\mathrm {d} z}{\mathrm {d} x}}={\frac {\mathrm {d} z}{\mathrm {d} y}}\,{\frac {\mathrm {d} y}{\mathrm {d} x}},\,$ which is easy to remember, because it as if dy in the numerator and the denominator of the right hand side cancels.

## Multivariable calculus

The extension of the chain rule to multivariable functions may be achieved by considering the derivative as a linear approximation to a differentiable function.

Now let $F:\mathbf {R} ^{n}\rightarrow \mathbf {R} ^{m}$ and $G:\mathbf {R} ^{m}\rightarrow \mathbf {R} ^{p}$ be functions with F having derivative $\mathrm {D} F$ at $a\in \mathbf {R} ^{n}$ and G having derivative $\mathrm {D} G$ at $F(a)\in \mathbf {R} ^{m}$ . Thus $\mathrm {D} F$ is a linear map from $\mathbf {R} ^{n}\rightarrow \mathbf {R} ^{m}$ and $\mathrm {D} G$ is a linear map from $\mathbf {R} ^{m}\rightarrow \mathbf {R} ^{p}$ . Then $F\circ G$ is differentiable at $a\in \mathbf {R} ^{n}$ with derivative

$\mathrm {D} (F\circ G)=\mathrm {D} F\circ \mathrm {D} G.\,$ 