Derivatives of Trig Functions

Question

For f(x) = e^sin(x) use your graphing calculator to find the number of zeros for f '(x) on the closed interval [0, 2π].

Options:

1. 1
2. 2
3. 3
4. 4

Answer 1

You could use a graphing calculator to find this answer,

but assuming you can do that and want to know how to verify the number of zeros analytically:

if f(x)=e^(sin x)

the derivative with respect to x of e^u where u can be a constant or a function of another variable such as x is: (e^(u))*du/dx by the chain rule

thus the derivative of e^(sin x)

is: f’(x)=(e^(sin x))*cos(x)

sin x goes from a minimum value of -1 to a maximum value of 1 for x on the range of 0 to 2pi

and e^sin(x) thus never goes to 0 on this range from 0 to 2pi

so f’(x) is only zero when cos(x) = 0 on the range of 0 to 2pi

by the unit circle from trigonometry:

cos(pi/2)=0

cos(3pi/2)=0

so the answer is 2