The Chain Rule: Learn It 4

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The Chain Rule: Learn It 4

The Chain Rule Using Leibniz’s Notation

As with other derivatives that we have seen, we can express the chain rule using Leibniz’s notation. This notation for the chain rule is used heavily in physics applications.

For $h(x)=f(g(x))$ , let $u=g(x)$ and $y=h(x)=g(u)$ . Thus,

$h^{\prime}(x)=\frac{dy}{dx}, \, f^{\prime}(g(x))=f^{\prime}(u)=\frac{dy}{du}$ , and

$g^{\prime}(x)=\frac{du}{dx}$

Consequently,

$\frac{dy}{dx}=h^{\prime}(x)=f^{\prime}(g(x))g^{\prime}(x)=\frac{dy}{du} \cdot \frac{du}{dx}$

chain rule using Leibniz’s notation

If $y$ is a function of $u$ , and $u$ is a function of $x$ , then

$\dfrac{dy}{dx}=\dfrac{dy}{du} \cdot \dfrac{du}{dx}$

Find the derivative of $y=\left(\dfrac{x}{3x+2}\right)^5$

First, let $u=\frac{x}{3x+2}$ . Thus, $y=u^5$ . Next, find $\frac{du}{dx}$ and $\frac{dy}{du}$ .

Using the quotient rule,

$\frac{du}{dx}=\frac{2}{(3x+2)^2}$

and

$\frac{dy}{du}=5u^4$

Finally, we put it all together.

$\begin{array}{lllll}\frac{dy}{dx} & =\frac{dy}{du} \cdot \frac{du}{dx} & & & \text{Apply the chain rule.} \\ & =5u^4 \cdot \frac{2}{(3x+2)^2} & & & \text{Substitute} \, \frac{dy}{du}=5u^4 \, \text{and} \, \frac{du}{dx}=\frac{2}{(3x+2)^2}. \\ & =5(\frac{x}{3x+2})^4 \cdot \frac{2}{(3x+2)^2} & & & \text{Substitute} \, u=\frac{x}{3x+2}. \\ & =\frac{10x^4}{(3x+2)^6} & & & \text{Simplify.} \end{array}$

It is important to remember that, when using the Leibniz form of the chain rule, the final answer must be expressed entirely in terms of the original variable given in the problem.

Find the derivative of $y= \tan (4x^2-3x+1)$

First, let $u=4x^2-3x+1$ . Then $y= \tan u$ . Next, find $\frac{du}{dx}$ and $\frac{dy}{du}$ :

$\frac{du}{dx}=8x-3$ and

$\frac{dy}{du}=\sec^2 u$ .

Finally, we put it all together.

$\begin{array}{lllll}\frac{dy}{dx} & =\frac{dy}{du} \cdot \frac{du}{dx} & & & \text{Apply the chain rule.} \\ & = \sec^2 u \cdot (8x-3) & & & \text{Use} \, \frac{du}{dx}=8x-3 \, \text{and} \, \frac{dy}{du}= \sec^2 u. \\ & = \sec^2 (4x^2-3x+1) \cdot (8x-3) & & & \text{Substitute} \, u=4x^2-3x+1. \end{array}$