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Topics in Calculus
Differential calculus
	Derivative; Change of variables; Implicit differentiation; Taylor's theorem; Related rates; Identities; Rules:; Power rule, Product rule, Quotient rule, Chain rule
Integral calculus
	Integral; Lists of integrals; Improper integrals; Integration by:; parts, disks, cylindrical; shells, substitution,; trigonometric substitution,; partial fractions, changing order
Vector calculus
	Gradient; Divergence; Curl; Laplacian; Gradient theorem; Green's theorem; Stokes' theorem; Divergence theorem
Multivariable calculus
	Matrix calculus; Partial derivative; Multiple integral; Line integral; Surface integral; Volume integral; Jacobian

For other uses, see Chain rule (disambiguation).

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. Schematically,

$\frac {\mathrm dy}{\mathrm dx} = \frac {\mathrm dy} {\mathrm du} \cdot\frac {\mathrm du}{\mathrm dx}.\,\!$

[edit] Informal discussion

For an explanation of notation used in this section, see Function composition.

The chain rule states that, under appropriate conditions,

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

which in short form is written as

$(f \circ g)' = (f'\circ g) \cdot g'\,\!$ .

Alternatively, in the Leibniz notation, the chain rule is

$\frac {\mathrm dy}{\mathrm dx} = \frac {\mathrm dy} {\mathrm du} \cdot\frac {\mathrm du}{\mathrm dx}.\,\!$

The chain rule can be applied to as many composed functions as needed:

$(f_1 \circ f_2 \circ f_3 \circ \cdots \circ f_n)'(x) = f'_1 \circ f_2 \circ f_3 \circ \cdots \circ f_n(x) \cdot f'_2 \circ f_3 \circ \cdots \circ f_n(x) \cdot f'_3 \circ \cdots \circ f_n(x) \cdot \,\cdots\, \cdot f'_n(x)$ .

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

[edit] Theorem

The chain rule in one variable may be stated more completely as follows.^[1] Let g be a real-valued function on (a,b) which is differentiable at c ∈ (a,b); and suppose that f is a real-valued function defined on an interval I containing the range of g and suppose further that g(c) is an interior point of I. If f is differentiable at g(c), then

$(f\circ g)(x)$ is differentiable at x = c, and
$(f\circ g)'(c) = f'(g(c))g'(c).$

[edit] Examples

[edit] Example I

Suppose that a mountain climber ascends at a rate of 0.5 kilometers per hour. The temperature is lower at higher elevations; suppose the rate by which it decreases is 6 °C per kilometer. To calculate the decrease in air temperature per unit time that the climber experiences, one multiplies 6 °C per kilometer by 0.5 kilometer per hour, to obtain 3 °C per hour. This calculation is a typical chain rule application.

[edit] Example II

Consider the function f(x) = (x² + 1)³. Since f(x) = h(g(x)) where g(x) = U = x² + 1 and h(U) = U³ it follows from the chain rule that

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

In order to differentiate the trigonometric function

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

one can write f(x) = h(g(x)) with h(x) = sin x 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).

$\frac{\mathrm d}{\mathrm dx}\arctan x = \frac{1}{1+x^2} \,\!$

Thus, by the chain rule,

$\frac{\mathrm d}{\mathrm dx}\arctan f(x) = \frac{f'(x)}{1+f^2(x)}\,\! ,$

and in particular,

$\frac{\mathrm d}{\mathrm dx}\arctan(\sin x) = \frac{\cos x}{1+\sin^2 x}\,\! .$

[edit] Chain rule for several variables

The chain rule works for functions of more than one variable.^[2] 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

${\ \mathrm dz \over \mathrm dt}={\partial z \over \partial x}{\mathrm dx \over \mathrm dt}+{\partial z \over \partial y}{\mathrm dy \over \mathrm dt}\,\!$ .

Suppose that each argument 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 consider

$\vec r = (u,v)$

above as a Cartesian 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 $\vec r$ :

$\frac{\partial z}{\partial x}=\vec \nabla f(u,v) \cdot \frac{\partial \vec r}{\partial x}.$

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] Example I

Given $u = x 2 + 2 y$ where $x = r sin(t)$ and $y = sin 2 (t)$ , determine the value of ${\partial u \over \partial r}$ and ${\partial u \over \partial t}$ using the chain rule.

${\partial u \over \partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r} = \left(2x\right)\left(\sin(t)\right)+\left(2\right)\left(0\right)=2r\sin^2(t)$

and

${\partial u \over \partial t}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t} = \left(2x\right)\left(r\cos(t)\right)+\left(2\right)\left(2\sin(t)\cos(t)\right)=$

$2\left(r\sin(t)\right)r\cos(t)+4\sin(t)\cos(t) = 2\left(r^2+2\right)\sin(t)\cos(t).$

[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 ε(δ) → 0 as δ → 0. Similarly,

$f(g(x)+\alpha) - f(g(x)) = \alpha f'(g(x)) + \eta(\alpha)\alpha \,\!$

where η(α) → 0 as α → 0. Define also^[3] that

$\eta(0) = 0 \,\!$

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 δ → 0, α_δ / δ → g′(x) and α_δ → 0, and thus η(α_δ) → 0. It follows that

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

To prove the multivariate chain rule, we will deal with the case of functions of two variables; a similar proof can be constructed for functions of three or more variables. Let x(t), y(t) be differentiable functions of t and assume f(x, y) has a gradient. If we set $Δ x = x (t + h) - x (t)$ and $Δ y = y (t + h) - y (t)$ , then we have:

$f'(x(t), y(t)) = \lim_{h\rightarrow 0} \frac{f(x(t + h), y(t + h)) - f(x(t), y(t))}{h}\,\!$

$= \lim_{h\rightarrow 0} \frac{f(x + \Delta x, y + \Delta y) - f(x, y + \Delta y) + f(x, y + \Delta y) - f(x, y)}{h}\,\!$

$= \lim_{h\rightarrow 0} \frac{f(x + \Delta x, y + \Delta y) - f(x, y + \Delta y)}{h} + \lim_{h\rightarrow 0} \frac{f(x, y + \Delta y) - f(x, y)}{h}\,\!$ .

When x is constant, we can regard $f (x, y)$ as a function $f x (y)$ of $y$ . Thus the limit on the right is equal to the derivative of $f x (y (t))$ , which by the single variable chain rule is $\frac{\partial f}{\partial y} \frac{\mathrm dy}{\mathrm dt}\,\!$ .

To calculate the limit on the left, regard $f (x, y + Δ y)$ as a function $f y + Δ y (x)$ of $x$ . By the mean value theorem, we can select a real number $c \in [x, x + \Delta x]$ such that the numerator on the left limit is equal to $\Delta x \frac{\mathrm df_{y + \Delta y}}{\mathrm dx}(c)\,\!$ . So the left limit is equal to $\lim_{h\rightarrow 0}\frac{\Delta x}{h}\frac{\mathrm df_{y + \Delta y}}{\mathrm dx}(c)$ , which equals $\frac{\partial f}{\partial x} \frac{\mathrm dx}{dt}.\,\!$

Thus, it follows that

$f'(x(t), y(t)) = \frac{\partial f}{\partial x}\frac{\mathrm dx}{\mathrm dt} + \frac{\partial f}{\partial y}\frac{\mathrm dy}{\mathrm dt} = \nabla f(x, y) \cdot (x', y')\,\!$ .

[edit] The fundamental chain rule

The chain rule is a fundamental property of all definitions of derivatives 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 C^k 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

$\frac{\mathrm d (f \circ g) }{\mathrm dx} = \frac{\mathrm df}{\mathrm dg}\frac{\mathrm dg}{\mathrm dx}$

$\frac{\mathrm d^2 (f \circ g) }{\mathrm d x^2} = \frac{\mathrm d^2 f}{\mathrm d g^2}\left(\frac{\mathrm dg}{\mathrm dx}\right)^2 + \frac{\mathrm df}{\mathrm dg}\frac{\mathrm d^2 g}{\mathrm dx^2}$

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

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

[edit] See also

[edit] References

^ Apostol, Tom (1974). Mathematical analysis (2nd ed. ed.). Addison Wesley. Theorem 5.5.
^ "The Multivariable Chain Rule". http://www.math.hmc.edu/calculus/tutorials/multichainrule/. Retrieved 2009-11-06.
^ To see that this is needed, suppose for example that g is a constant function.

[edit] External links

http://www.brightstorm.com/d/math/s/calculus/u/techniques-of-differentiation/t/the-chain-rule | Brightstorm

Weisstein, Eric W., "Chain Rule" from MathWorld.

[0] Apostol, Tom (1974). Mathematical analysis (2nd ed. ed.). Addison Wesley. Theorem 5.5.

[1] "The Multivariable Chain Rule". http://www.math.hmc.edu/calculus/tutorials/multichainrule/. Retrieved 2009-11-06.

[2] To see that this is needed, suppose for example that g is a constant function.

[1]

[2]

[3]

Contents

[edit] Informal discussion

[edit] Theorem

[edit] Examples

[edit] Example I

[edit] Example II

[edit] Example III

[edit] Chain rule for several variables

[edit] Example I

[edit] Proof of the chain rule

[edit] The fundamental chain rule

[edit] Tensors and the chain rule

[edit] Higher derivatives

[edit] See also

[edit] References

[edit] External links