Chapter 3.4, Problem 47E

(a)

To determine

To find: The derivative of $f\left(x\right)=x\sqrt{2-x^2}$.

Answer to Problem 47E

The derivative of $f\left(x\right)=x\sqrt{2-x^2}$ is $\underset{\_}{f^{\prime }\left(x\right)=\frac{2-2x^2}{\sqrt{2-x^2}}}$.

Explanation of Solution

Given:

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

Result used:

The Power Rule combined with the Chain Rule:

If n is any real number and $g\left(x\right)$ is differentiable function, then

$\frac{d}{dx}{\left[g\left(x\right)\right]}^n=n{\left[g\left(x\right)\right]}^{n-1}{g}^{\prime }\left(x\right)$ (1)

Derivative Rule: Product Rule:

If $f\left(x\right)$. and $g\left(x\right)$ are both differentiable function, then

$\frac{d}{dx}\left[f\left(x\right)g\left(x\right)\right]=f\left(x\right)\frac{d}{dx}\left[g\left(x\right)\right]+g\left(x\right)\frac{d}{dx}\left[f\left(x\right)\right]$ (2)

Calculation:

Obtain the derivative of $f\left(x\right)$.

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

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

$\begin{array}{c}{f}^{\prime }\left(x\right)=x\frac{d}{dx}\left[\sqrt{2-x^2}\right]+\sqrt{2-x^2}\frac{d}{dx}\left[x\right]\\ =x\frac{d}{dx}{\left(2-x^2\right)}^{\frac{1}{2}}+\sqrt{2-x^2}\left[1x^{1-1}\right]\\ =x\frac{d}{dx}{\left(2-x^2\right)}^{\frac{1}{2}}+\sqrt{2-x^2}\left[x^0\right]\\ =x\frac{d}{dx}{\left(2-x^2\right)}^{\frac{1}{2}}+\sqrt{2-x^2}\end{array}$

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

$\begin{array}{c}{f}^{\prime }\left(x\right)=x\frac{1}{2}{\left(2-x^2\right)}^{\frac{1}{2}-1}\frac{d}{dx}\left(2-x^2\right)+\sqrt{2-x^2}\\ =x\frac{1}{2}{\left(2-x^2\right)}^{\frac{1-2}{2}}\left(\frac{d}{dx}\left(2\right)-\frac{d}{dx}\left(x^2\right)\right)+\sqrt{2-x^2}\\ =x\frac{1}{2}{\left(2-x^2\right)}^{\frac{-1}{2}}\left(0-\left(2x^{2-1}\right)\right)+\sqrt{2-x^2}\\ =x\frac{1}{2{\left(2-x^2\right)}^{\frac{1}{2}}}\left(-2x^1\right)+\sqrt{2-x^2}\end{array}$

Simplify further and obtain the derivative,

$\begin{array}{c}{f}^{\prime }\left(x\right)=-\frac{x^2}{\sqrt{2-x^2}}+\sqrt{2-x^2}\\ =\frac{-x^2+\sqrt{2-x^2}\sqrt{2-x^2}}{\sqrt{2-x^2}}\\ =\frac{-x^2+2-x^2}{\sqrt{2-x^2}}\\ =\frac{2-2x^2}{\sqrt{2-x^2}}\end{array}$

Therefore, the derivative of $f\left(x\right)=x\sqrt{2-x^2}$ is $\underset{\_}{f^{\prime }\left(x\right)=\frac{2\left(1-x^2\right)}{\sqrt{2-x^2}}}$.

(b)

To determine

To sketch: The graph of $f\left(x\right)$ and $f^{\prime }\left(x\right)$.

Explanation of Solution

Given:

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

The derivative of $f\left(x\right)$ is $\frac{2-2x^2}{\sqrt{2-x^2}}$.

Graph:

Use the online graphing calculator and draw the graph of the functions as shown below in Figure 1.

From Figure 1 it is observe that,

If $f^{\prime }\left(x\right)$ is positive then $f\left(x\right)$ is increasing function.

If $f^{\prime }\left(x\right)$ is negative then $f\left(x\right)$ is decreasing function.

If $f\left(x\right)$ is local extrema (that is local minimum or local maximum), then $f^{\prime }\left(x\right)$ is crosses the x axis $\left(f^{\prime }\left(x\right)=0\right)$.

Hence, the function $f^{\prime }\left(x\right)$ is reasonable.