27. (a) If $f(x)=\sec x-x$, find $f^{\prime}(x)$. (b) Check to see that your answer to part (a) is reasonable by graphing both $f$ and $f^{\prime}$ for $|x|<\pi / 2$.

27. (a) If $f(x)=\sec x-x$, find $f^{\prime}(x)$.

(b) Check to see that your answer to part (a) is reasonable by graphing both $f$ and $f^{\prime}$ for $|x|<\pi / 2$.

To find: The derivative of the function f(x)=secx−x.

The derivative of f(x) is f′(x)=secxtanx−1_.

Given:

The function is f(x)=secx−x.

Derivative rules:

(1) Difference Rule: ddx[f(x)−g(x)]=ddx[f(x)]−ddx[g(x)]

(2) Power Rule: ddx(xn)=nxn−1

Calculation:

Obtain the derivative of f(x).

f′(x)=ddx(f(x)) =ddx(secx−x)

Apply the difference rule (1),

f′(x)=ddx[secx]−ddx[x]=secxtanx−ddx[x]

Apply the power rule (2) and simplify further,

f′(x)=secxtanx−[1x1−1]=secxtanx−1

Therefore, the derivative of f(x) is f′(x)=secxtanx−1_.

To check: The derivatives of f(x) obtained in part (a) is reasonable or not by using the graphs of f(x) and f′(x).

The derivatives of f(x) is reasonable.

Graph:

Use the online graphing calculator and draw the graph of f(x)=secx−x and f′(x)=secxtanx−1 as shown below in Figure 1.

Student Solutions Manual, Chapters 1-11 for Stewart's Single Variable Calculus, 8th (James Stewart Calculus), Chapter 2.4, Problem 27E

Observation:

From Figure 1, it is noticed that,

If f′(x) is positive then f(x) is increasing function.

If f′(x) is negative then f(x) is decreasing function.

If f′(x) crosses the x axis (f′(x)=0), then f(x) is local extrema (that is, local minimum or local maximum).

Therefore, it can be concluded that the derivative of the function f(x)=secx−x is reasonable.