Chapter 3.4, Problem 48E

(a)

To determine

To sketch: The slope of the function $f\left(x\right)$ by using the graph of $f\left(x\right)$.

Explanation of Solution

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 $f^{\prime }\left(x\right)$ as shown in Figure 1.

Graph:

The rough sketch of the derivative function, $f^{\prime }\left(x\right)$ by using the graph of the function $f\left(x\right)$ is shown below in Figure 1.

(b)

To determine

To calculate: The derivative of function $f\left(x\right)$.

Answer to Problem 48E

The derivative of $f\left(x\right)=\sin \left(x+\sin 2x\right)$ is $\underset{\_}{f^{\prime }\left(x\right)=\cos \left(x+\sin 2x\right)\left(1+2\cos 2x\right)}$.

Explanation of Solution

Result used: 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)

Calculation:

Obtain the derivative.

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

Let $h\left(x\right)=x+\sin 2x$ and $g\left(u\right)=\sin u\text{ where }u=h\left(x\right)$.

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

$f^{\prime }\left(x\right)=g^{\prime }\left(h\left(x\right)\right)\cdot h^{\prime }\left(x\right)$ (2)

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

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

Substitute $u=x+\sin 2x$ in the above equation,

$g^{\prime }\left(h\left(x\right)\right)=\cos \left(x+\sin 2x\right)$

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

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

$\begin{array}{c}{h}^{\prime }\left(x\right)=\frac{d}{dx}\left(x+\sin 2x\right)\\ =\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\sin 2x\right)\\ =\left(1x^{1-1}\right)+\cos 2x\cdot 2\\ =1+2\cos 2x\end{array}$

Thus the derivative is $h^{\prime }\left(x\right)=1+2\cos 2x$.

Substitute $\cos \left(x+\sin 2x\right)$ for $g^{\prime }\left(h\left(x\right)\right)$ and $1+2\cos 2x$ for $h^{\prime }\left(x\right)$ in equation (2),

$\begin{array}{c}{F}^{\prime }\left(x\right)=\cos \left(x+\sin 2x\right)\cdot 1+2\cos 2x\\ =\left(1+2\cos 2x\right)\cdot \cos \left(x+\sin 2x\right)\end{array}$

Therefore, the derivative of $f\left(x\right)=\sin \left(x+\sin 2x\right)$ is $\left(1+2\cos 2x\right)\cdot \cos \left(x+\sin 2x\right)$.

Graph:

Use the online graphing calculator to draw the graph of the derivative function $f^{\prime }\left(x\right)$ as shown below in Figure 2.

From Figure 2, it is observed that the given trigonometric function is differentiable everywhere and the rough sketch of the derivative $f^{\prime }\left(x\right)$ in Figure 1 is same as the graph of the derivative $f^{\prime }\left(x\right)$ in Figure 2.