Let f(x) = cx + ln(cos(x)). For what value of c is f '(π/4) = 5?

Question

Let f(x) = cx + ln(cos(x)). For what value of c is f '(π/4) = 5?

c=______

Answer 1

1. Find f'(x)
2. f'(x) = c - sin(x)/cos(x)
3. f'(x) = c - tan(x)
4. Plug in Values
5. 5 = c - tan(pi/4)
6. Solve
7. 5 + 1 = c
8. Answer
9. c = 6

Answer 2

The first step is to take the derivative of the function.

f (x) = cx + ln(cos(x)); then

f '(x) = c + 1/cos(x) * -sin (x)

f '(x) = c - tan(x)

I did this by following the chain rule. Derivative of the outside times the derivative of the inside.

So to find the derivative of ln(cos(x)) you first take the derivative of ln(u) where u = cos(x)

this is 1/u or 1/cos(x)

You then multiply by the derivative of u or cos(x)

this is -sin(x)

So by combining and simplifying the entire derivative becomes

f '(x) = c - tan(x)



In this case you are finding c when f '(π/4) = 5

Set the equation

f '(π/4) = c - tan(π/4) = 5 Remember tan(π/4) = 1

c - 1 = 5

c = 6