Chapter 3.3, Problem 26E

(a)

To determine

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

Answer to Problem 26E

The first derivative of $f\left(x\right)=e^x\cos x$ is $\underset{\_}{f^{\prime }\left(x\right)=e^x\left(\cos x-\sin x\right)}$.

The second derivative of $f\left(x\right)=e^x\cos x$ is $\underset{\_}{f^{″}\left(x\right)=-2e^x\sin x}$.

Explanation of Solution

Given:

The function is $f\left(x\right)=e^x\cos x$.

Derivative rules:

(1) Product Rule: $\frac{d}{dx}\left[f\left(x\right)g\left(x\right)\right]=f\left(x\right)\frac{d}{dx}\left[g\left(x\right)\right]+g\left(x\right)\frac{d}{dx}\left[f\left(x\right)\right]$

(2) Difference Rule: $\frac{d}{dx}\left[f\left(x\right)-g\left(x\right)\right]=\frac{d}{dx}\left[f\left(x\right)\right]-\frac{d}{dx}\left[g\left(x\right)\right]$

Calculation:

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

$\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(f\left(x\right)\right)\\ \text{=}\frac{d}{dx}\left(e^x\cos x\right)\end{array}$

Apply the product rule (1) and simplify further,

$\begin{array}{c}{f}^{\prime }\left(x\right)=e^x\frac{d}{dx}\left[\cos x\right]+\cos x\frac{d}{dx}\left[e^x\right]\\ \text{=}{e}^x\left[-\sin x\right]+\cos x\left[e^x\right]\\ \text{=}-e^x\sin x+e^x\cos x\\ \text{=}{e}^x\left(-\sin x+\cos x\right)\end{array}$

Therefore, the first derivative of $f\left(x\right)=e^x\cos x$ is $\underset{\_}{f^{\prime }\left(x\right)=e^x\left(\cos x-\sin x\right)}$.

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

Apply the difference rule (2),

$\begin{array}{c}{f}^{″}\left(x\right)=\frac{d}{dx}\left(f^{\prime }\left(x\right)\right)\\ =\frac{d}{dx}\left(e^x\left(-\sin x+\cos x\right)\right)\\ =\frac{d}{dx}\left(\left(e^x\cos x-e^x\sin x\right)\right)\\ =\left(\frac{d}{dx}\left(e^x\cos x\right)-\frac{d}{dx}\left(e^x\sin x\right)\right)\end{array}$

Apply the product rule (1),

$\begin{array}{c}{f}^{″}\left(x\right)=\left(e^x\frac{d}{dx}\left(\cos x\right)+\cos x\frac{d}{dx}\left(e^x\right)\right)-\left(e^x\frac{d}{dx}\left(\sin x\right)+\sin x\frac{d}{dx}\left(e^x\right)\right)\\ =\left(e^x\left(-\sin x\right)+\cos x\left(e^x\right)\right)-\left(e^x\left(\cos x\right)+\sin x\left(e^x\right)\right)\\ =-e^x\sin x+e^x\cos x-e^x\cos x-e^x\sin x\\ =-2e^x\sin x\end{array}$

Therefore, the derivative of $f^{\prime }\left(x\right)=e^x\left(-\sin x+\cos x\right)$ is $\underset{\_}{-2e^x\sin x}$.

(b)

To determine

To check: The derivatives $f\left(x\right)$, $f^{\prime }\left(x\right)$. and $f^{″}\left(x\right)$ are reasonable by comparing the graphs of $f\left(x\right)$, $f^{\prime }\left(x\right)$. and $f^{″}\left(x\right)$.

Answer to Problem 26E

The derivatives are reasonable.

Explanation of Solution

From part (a), the derivatives of $f\left(x\right)=e^x\cos x$ are, $f^{\prime }\left(x\right)=e^x\left(\cos x-\sin x\right)$ and $f^{″}\left(x\right)=-2e^x\sin x$.

Graph:

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

Form Figure 1, it is observed that the function $f\left(x\right)$ is continuous and monotonically decreasing on the positive real number.

The value of the derivative of the function at any point x can be estimated by drawing the tangent line at any point $\left(x,f\left(x\right)\right)$ and then obtain the slope.

Mark the slope of the tangent as a point in y-axis and the value of x as a point in x-axis in the graph of $f^{\prime }\left(x\right)$.

Proceed in the similar way at several points and obtain the rough graph of the $f^{\prime }\left(x\right)$.

Hence, it can be concluded that the derivative of the function $f^{\prime }\left(x\right)$ is reasonable.

The value of the derivative of the function at any point x can be estimated by drawing the tangent line at any point $\left(x,f^{\prime }\left(x\right)\right)$ and then obtain the slope.

Mark the slope of the tangent as a point in y-axis and the value of x as a point in x-axis in the graph of ${f^{\prime }}^{\prime }\left(x\right)$.

Proceed in the similar way at several points and obtain the rough graph of the ${f^{\prime }}^{\prime }\left(x\right)$.

Hence, it can be concluded that the derivative of the function ${f^{\prime }}^{\prime }\left(x\right)$ is reasonable.