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^{'} (x) = \frac{d y}{d x}, f^{'} (g (x)) = f^{'} (u) = \frac{d y}{d u}

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

g^{'} (x) = \frac{d u}{d x}

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

Consequently,

\frac{d y}{d x} = h^{'} (x) = f^{'} (g (x)) g^{'} (x) = \frac{d y}{d u} \cdot \frac{d u}{d x}

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

Chain Rule Using Leibniz’s Notation

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

\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}

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

Example: Taking a Derivative Using Leibniz’s Notation, 1

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

Show Solution

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{d u}{d x} = \frac{2}{(3 x + 2)^{2}}

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

and

\frac{d y}{d u} = 5 u^{4}

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

Finally, we put it all together.

\begin{array}{lllll} \frac{d y}{d x} & = \frac{d y}{d u} \cdot \frac{d u}{d x} & Apply the chain rule. \\ = 5 u^{4} \cdot \frac{2}{(3 x + 2)^{2}} & Substitute \frac{d y}{d u} = 5 u^{4} and \frac{d u}{d x} = \frac{2}{(3 x + 2)^{2}} . \\ = 5 (\frac{x}{3 x + 2})^{4} \cdot \frac{2}{(3 x + 2)^{2}} & Substitute u = \frac{x}{3 x + 2} . \\ = \frac{10 x^{4}}{(3 x + 2)^{6}} & Simplify. \end{array}

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

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.

Example: Taking a Derivative Using Leibniz’s Notation, 2

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

Show Solution

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

\frac{d u}{d x} = 8 x - 3

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

\frac{d y}{d u} = \sec^{2} u

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

Finally, we put it all together.

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

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

Try It

Use Leibniz’s notation to find the derivative of $y= \cos (x^3)$ . Make sure that the final answer is expressed entirely in terms of the variable $x$ .

Hint

Let $u=x^3$ .

Show Solution

$\frac{dy}{dx}=-3x^2 \sin (x^3)$

Watch the following video to see the worked solution to the above Try It.

Closed Captioning and Transcript Information for Video

Module 3: Derivatives