Chapter 3.6, Problem 35E

To determine

To calculate: The derivative of the function $f\left(x\right)=\sqrt{1-x^2}\arcsin \left(x\right)$ . Also sketch the graph of $f\text{ and }f\text{'}$ to compare the result.

Answer to Problem 35E

The derivative of the function is $f\text{'}\left(x\right)=1-\arcsin \left(x\right)\frac{x}{\sqrt{1-x^2}}$ .The blue curve depicts the graph of $f\text{'}\left(x\right)=1-\arcsin \left(x\right)\frac{x}{\sqrt{1-x^2}}$ and red curve depicts the graph of $f\left(x\right)=\sqrt{1-x^2}\arcsin \left(x\right)$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=\sqrt{1-x^2}\arcsin \left(x\right)$ .

Formula used:

Thechain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Power rule for differentiation is $\frac{d}{dx}{x}^n=n{x}^{n-1}$ .

Product rule for differentiation is $\frac{d}{dx}\left(f\cdot g\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)$ where f and g are functions of x .

Derivative of inverse sine function is $\frac{d}{dx}\arcsin \left(x\right)=\frac{1}{\sqrt{1-x^2}}$ .

Calculation:

Consider the function $f\left(x\right)=\sqrt{1-x^2}\arcsin \left(x\right)$ .

Differentiate both sides with respect to x ,

  $\begin{array}{c}\frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(\sqrt{1-x^2}\arcsin \left(x\right)\right)\\ f\text{'}\left(x\right)=\frac{d}{dx}\left(\sqrt{1-x^2}\arcsin \left(x\right)\right)\end{array}$

Recall that power rule for differentiation is $\frac{d}{dx}{x}^n=n{x}^{n-1}$ and chain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Also for the terms of the above expression, apply the product rule for differentiation.

Recall that product rule for differentiation is $\frac{d}{dx}\left(f\cdot g\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)$ where f and g are functions of x .Derivative of inverse sine function is $\frac{d}{dx}\arcsin \left(x\right)=\frac{1}{\sqrt{1-x^2}}$ .

Apply it.

  $\begin{array}{c}f\text{'}\left(x\right)=\sqrt{1-x^2}\frac{d}{dx}\left(\arcsin \left(x\right)\right)+\arcsin \left(x\right)\frac{d}{dx}\left(\sqrt{1-x^2}\right)\\ =\sqrt{1-x^2}\frac{1}{\sqrt{1-x^2}}+\arcsin \left(x\right)\frac{1}{2\sqrt{1-x^2}}\frac{d}{dx}\left(1-x^2\right)\\ =\sqrt{1-x^2}\frac{1}{\sqrt{1-x^2}}+\arcsin \left(x\right)\frac{1}{2\sqrt{1-x^2}}\left(-2x\right)\\ =1-\arcsin \left(x\right)\frac{x}{\sqrt{1-x^2}}\end{array}$

Thus, the derivative of the function is $f\text{'}\left(x\right)=1-\arcsin \left(x\right)\frac{x}{\sqrt{1-x^2}}$ .

Now, use a graphing window to plot the function and its derivative. Sketch the graph of $f\left(x\right)=\sqrt{1-x^2}\arcsin \left(x\right)$ and $f\text{'}\left(x\right)=1-\arcsin \left(x\right)\frac{x}{\sqrt{1-x^2}}$ .

The blue curve depicts the graph of $f\text{'}\left(x\right)=1-\arcsin \left(x\right)\frac{x}{\sqrt{1-x^2}}$ and red curve depicts the graph of $f\left(x\right)=\sqrt{1-x^2}\arcsin \left(x\right)$ . The graph of $f\text{'}\left(x\right)$ intersects the x -axis at the same values when the graph of the function $f\left(x\right)$ levels off. Also observe that the graph of $f\text{'}\left(x\right)$ attains the maximum value at the same value of x when the graph of the function $f\left(x\right)$ has the steepest positive incline.