Chain rule

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Topics in calculus

Fundamental theorem
Limits of functions
Continuity
Vector calculus
Tensor calculus
Mean value theorem
Differentiation

Product rule
Quotient rule
Chain rule
Implicit differentiation
Taylor's theorem
Related rates
Table of derivatives
Integration

Lists of integrals
Improper integrals
Integration by: parts, disks,
cylindrical shells, substitution,
trigonometric substitution

In calculus, the chain rule is a formula for the derivative of the composite of two functions.

In intuitive terms, if a variable, y, depends on a second variable, u, which in turn depends on a third variable, x, then the rate of change of y with respect to x can be computed as the rate of change of y with respect to u multiplied by the rate of change of u with respect to x.

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[edit] Definition

The chain rule states that

(f \circ g)'(x) = (f(g(x)))' = f'(g(x)) g'(x),\,

which in short form is written as (f \circ g)' = f'\circ g\cdot g'.

Alternatively, in the Leibniz notation, the chain rule is

\frac {df}{dx} = \frac {df} {dg} \frac {dg}{dx}.

In integration, the counterpart to the chain rule is the substitution rule.

[edit] Examples

[edit] Example I

Suppose, that one is climbing a mountain at a rate of 0.5 kilometres per hour. The temperature is lower at higher elevations; suppose the rate by which it decreases is 6 °C per kilometre. If one multiplies 6 °C per kilometre by 0.5 kilometre per hour, one obtains 3 °C per hour. This calculation is a typical chain rule application.

[edit] Example II

Consider f(x) = (x2 + 1)3. We have f(x) = h(g(x)) where g(x) = x2 + 1 and h(x) = x3. Thus,

f '(x) \, = 3(x^2 + 1)^2(2x) \,
= 6x(x^2 + 1)^2. \,

In order to differentiate the trigonometric function

f(x) = \sin(x^2),\,

one can write f(x) = h(g(x)) with h(x) = sinx and g(x) = x2. The chain rule then yields

f'(x) = 2x \cos(x^2) \,

since h'(g(x)) = cos(x2) and g'(x) = 2x.

[edit] Example III

Differentiate \arctan\,\sin\, x, etc.

\frac{d}{dx}\arctan\,x\,=\,\frac{1}{1+x^2}
\frac{d}{dx}\arctan\,f(x)\,=\,\frac{f'(x)}{1+f^2(x)}
\frac{d}{dx}\arctan\,\sin\,x\,=\,\frac{\cos\,x}{1+\sin^2\,x}

[edit] Chain rule for several variables

The chain rule works for functions of more than one variable. Consider the function z = f(x,y) where x = g(t) and y = h(t), and g(t) and h(t) are differentiable with respect to t, then

{\ dz \over dt}={\partial f \over \partial x}{dx \over dt}+{\partial f \over \partial y}{dy \over dt}

Suppose that each function of z = f(u,v) is a two-variable function such that u = h(x,y) and v = g(x,y), and that these functions are all differentiable. Then the chain rule would look like:

{\partial z \over \partial x}={\partial z \over \partial u}{\partial u \over \partial x}+{\partial z \over \partial v}{\partial v \over \partial x}


{\partial z \over \partial y}={\partial z \over \partial u}{\partial u \over \partial y}+{\partial z \over \partial v}{\partial v \over \partial y}

If we considered \vec r = (u,v) above as a vector function, we can use vector notation to write the above equivalently as the dot product of the gradient of f and a derivative of \vec r:

\frac{\partial f}{\partial x}=\vec \nabla f \cdot \frac{\partial \vec r}{\partial x}

More generally, for functions of vectors to vectors, the chain rule says that the Jacobian matrix of a composite function is the product of the Jacobian matrices of the two functions:

\frac{\partial(z_1,\ldots,z_m)}{\partial(x_1,\ldots,x_p)} = \frac{\partial(z_1,\ldots,z_m)}{\partial(y_1,\ldots,y_n)} \frac{\partial(y_1,\ldots,y_n)}{\partial(x_1,\ldots,x_p)}

[edit] Proof of the chain rule

Let f and g be functions and let x be a number such that f is differentiable at g(x) and g is differentiable at x. Then by the definition of differentiability,

g(x+\delta)-g(x)= \delta g'(x) + \epsilon(\delta)\delta \, where \epsilon(\delta) \to 0 \, as \delta\to 0.

Similarly,

f(g(x)+\alpha) - f(g(x)) = \alpha f'(g(x)) + \eta(\alpha)\alpha \, where \eta(\alpha) \to 0 \, as \alpha\to 0. \,

Now

f(g(x+\delta))-f(g(x))\, = f(g(x) + \delta g'(x)+\epsilon(\delta)\delta) - f(g(x)) \,
= \alpha_\delta f'(g(x)) + \eta(\alpha_\delta)\alpha_\delta \,

where \alpha_\delta = \delta g'(x) + \epsilon(\delta)\delta \,. Observe that as \delta\to 0, \frac{\alpha_\delta}{\delta}\to g'(x) and \alpha_\delta \to 0, thus \eta(\alpha_\delta)\to 0. Therefore

\frac{f(g(x+\delta))-f(g(x))}{\delta} \to g'(x)f'(g(x))\mbox{ as } \delta \to 0.

[edit] The fundamental chain rule

The chain rule is a fundamental property of all definitions of derivative and is therefore valid in much more general contexts. For instance, if E, F and G are Banach spaces (which includes Euclidean space) and f : E → F and g : F → G are functions, and if x is an element of E such that f is differentiable at x and g is differentiable at f(x), then the derivative (the Fréchet derivative) of the composition g o f at the point x is given by

\mbox{D}_x\left(g \circ f\right) = \mbox{D}_{f\left(x\right)}\left(g\right) \circ \mbox{D}_x\left(f\right).

Note that the derivatives here are linear maps and not numbers. If the linear maps are represented as matrices (namely Jacobians), the composition on the right hand side turns into a matrix multiplication.

A particularly clear formulation of the chain rule can be achieved in the most general setting: let M, N and P be Ck manifolds (or even Banach-manifolds) and let

f : M → N and g : N → P

be differentiable maps. The derivative of f, denoted by df, is then a map from the tangent bundle of M to the tangent bundle of N, and we may write

\mbox{d}\left(g \circ f\right) = \mbox{d}g \circ \mbox{d}f.

In this way, the formation of derivatives and tangent bundles is seen as a functor on the category of C∞ manifolds with C∞ maps as morphisms.

[edit] Tensors and the chain rule

See tensor field for an advanced explanation of the fundamental role the chain rule plays in the geometric nature of tensors.

[edit] Higher derivatives

Faà di Bruno's formula generalizes the chain rule to higher derivatives. The first few derivatives are

\frac{df}{dx} = \frac{df}{dg}\frac{dg}{dx}
\frac{d^2 f}{d x^2} = \frac{d^2 f}{d g^2}\left(\frac{dg}{dx}\right)^2 + \frac{df}{dg}\frac{d^2 g}{dx^2}
\frac{d^3 f}{d x^3} = \frac{d^3 f}{d g^3} \left(\frac{dg}{dx}\right)^3 + 3 \frac{d^2 f}{d g^2} \frac{dg}{dx} \frac{d^2 g}{d x^2} + \frac{df}{dg} \frac{d^3 g}{d x^3}
\frac{d^4 f}{d x^4} =\frac{d^4 f}{dg^4} \left(\frac{dg}{dx}\right)^4 + 6 \frac{d^3 f}{d g^3} \left(\frac{dg}{dx}\right)^2 \frac{d^2 g}{d x^2} + \frac{d^2 f}{d g^2} \left\{ 4 \frac{dg}{dx} \frac{d^3 g}{dx^3} + 3\left(\frac{d^2 g}{dx^2}\right)^2\right\} + \frac{df}{dg}\frac{d^4 g}{dx^4}

[edit] See also

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