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Concept Version 7
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The Chain Rule

The chain rule is a formula for computing the derivative of the composition of two or more functions.

Learning Objective

  • Calculate the derivative of a composition of functions using the chain rule


Key Points

    • If $f$ is a function and $g$ is a function, then the chain rule expresses the derivative of the composite function $f \circ g$ in terms of the derivatives of $f$ and $g$.
    • The chain rule can be applied sequentially for as many functions as are nested inside one another.
    • The chain rule for $f \circ g(x)$ is $\frac{df}{dx} = \frac{df}{dg}\frac{dg}{dx}$.

Term

  • composite

    a function of a function


Full Text

The chain rule is a formula for computing the derivative of the composition of two or more functions. That is, if $f$ is a function and $g$ is a function, then the chain rule expresses the derivative of the composite function $f \circ g$ in terms of the derivatives of $f$ and $g$.

For example, following the chain rule for $f \circ g(x) = f[g(x)]$ yields:

$\displaystyle{\frac{df}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx}}$

The method is called the "chain rule" because it can be applied sequentially to as many functions as are nested inside one another. For example, if $f$ is a function of $g$, which is in turn a function of $h$, which is in turn a function of $x$—that is, $f(g(h(x)))$—then the derivative of $f$ with respect to $x$ is:

$\displaystyle{\frac{df}{dx} = \frac{df}{dg} \cdot \frac{dg}{dh} \cdot \frac{dh}{dx}}$

The chain rule has broad applications in physics, chemistry, and engineering, as well as for the study of related rates in many other disciplines. The chain rule can also be generalized to multiple variables in cases where the nested functions depend on more than one variable.

Skydiving

The path of a skydiver relies on many variables such as time and height. Use of the chain rule is needed for the complicated calculation.

Consider the function $f(x) = (x^2 + 1)^3$. Using the chain rule yields:

$f(x) = (x^2 + 1)^3$

$u(x) = x^2 + 1$

$f(x) = [u(x)]^3$

Substituting $f(u)$ and $u(x)$, we get

$\displaystyle{\frac {df}{dx} = \frac {df}{du} \cdot \frac {du}{dx}}$

$\displaystyle{ \qquad = \frac {d}{du}u^3 \frac {d}{dx} (x^2 + 1)}$

$\quad \quad = (3u^2) (2x)$

$\quad \quad = \left ( 3(x^2 + 1)^2 \right ) \cdot (2x)$

$\quad \quad = 6x(x^2 + 1)^2$

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