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 [latex]h(x)=f(g(x))[/latex], let [latex]u=g(x)[/latex] and [latex]y=h(x)=g(u)[/latex]. Thus,
[latex]h^{\prime}(x)=\frac{dy}{dx}, \, f^{\prime}(g(x))=f^{\prime}(u)=\frac{dy}{du}[/latex], and [latex]g^{\prime}(x)=\frac{du}{dx}[/latex]
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 [latex]y= \tan (4x^2-3x+1)[/latex]
First, let [latex]u=4x^2-3x+1[/latex]. Then [latex]y= \tan u[/latex]. Next, find [latex]\frac{du}{dx}[/latex] and [latex]\frac{dy}{du}[/latex]:
[latex]\frac{du}{dx}=8x-3[/latex] and [latex]\frac{dy}{du}=\sec^2 u[/latex].