Chapter 3.4, Problem 3E

To determine

To find: The composite function in the form $f\left(g\left(x\right)\right)$ and obtain the derivative of y.

Answer to Problem 3E

The inner function is $u=\pi x$ and the outer function is $f\left(u\right)=\tan u$.

The derivative of y is $\underset{\_}{y^{\prime }\left(x\right)=\pi {\sec }^2\pi x}$.

Explanation of Solution

Given:

The function is $y=\tan \pi x$.

Formula used:

The Chain Rule:

If h is differentiable at x and g is differentiable at $h\left(x\right)$, then the composite function $F=g\circ h$ defined by $F\left(x\right)=g\left(h\left(x\right)\right)$ is differentiable at x and $F^{\prime }$ is given by the product

$F^{\prime }\left(x\right)=g^{\prime }\left(h\left(x\right)\right)\cdot h^{\prime }\left(x\right)$ (1)

Power Rule:

If n is positive integer, then $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ (2)

Calculation:

Let the inner function be $u=g\left(x\right)$ and the outer function be $y=f\left(u\right)$.

Then, $g\left(x\right)=\pi x$ and $f\left(u\right)=\tan u$. That is,

$\begin{array}{c}y=\tan \pi x\\ =f\left(\pi x\right)\\ =f\left(g\left(x\right)\right)\end{array}$

Therefore, $y=f\left(g\left(x\right)\right)$.

Hence, the inner function is $u=\pi x$ and the outer function is $f\left(u\right)=\tan u$.

Thus, the required form of composite function is $f\left(g\left(x\right)\right)=\tan u$.

Obtain the derivative of y.

Let $h\left(x\right)=\pi x$ and $g\left(u\right)=\tan u\text{ where }u=h\left(x\right)$.

Apply the chain rule as shown in equation (1),

$y^{\prime }\left(x\right)=g^{\prime }\left(h\left(x\right)\right)\cdot h^{\prime }\left(x\right)$ (3)

The derivative $g^{\prime }\left(h\left(x\right)\right)$ is computed as follows,

$\begin{array}{c}{g}^{\prime }\left(h\left(x\right)\right)=g^{\prime }\left(u\right)\\ =\frac{d}{du}\left(g\left(u\right)\right)\\ =\frac{d}{du}\left(\tan u\right)\\ ={\sec }^{2}u\end{array}$

Substitute $u=\pi x$ in the above equation,

$g^{\prime }\left(h\left(x\right)\right)={\sec }^2\pi x$

Thus, the derivative $g^{\prime }\left(h\left(x\right)\right)$ is $g^{\prime }\left(h\left(x\right)\right)={\sec }^2\pi x$.

The derivative of $h\left(x\right)$ is computed as follows,

$h^{\prime }\left(x\right)=\frac{d}{dx}\left(\pi x\right)$

Apply the power rule as shown in equation (2),

$\begin{array}{c}{h}^{\prime }\left(x\right)=\pi \left[1x^{1-1}\right]\\ =\pi \left(1x^0\right)\\ =\pi \left(1\right)\\ =\pi \end{array}$

Substitute ${\sec }^2\pi x$ for $g^{\prime }\left(h\left(x\right)\right)$ and $\pi$ for $h^{\prime }\left(x\right)$ in equation (3),

$\begin{array}{c}{g}^{\prime }\left(h\left(x\right)\right)\cdot h^{\prime }\left(x\right)={\sec }^2\pi x\left(\pi \right)\\ =\pi {\sec }^2\pi x\end{array}$

Therefore, the derivative of $y=\tan \pi x$ is $\underset{\_}{y^{\prime }\left(x\right)=\pi {\sec }^2\pi x}$.