# Differentiate f(x) = sec(5x)*ln(cos(5x))

### 3 Answers | Add Yours

The function f(x) = `sec(5x)*ln(cos(5x))`

To find the derivative use the product rule and the chain rule.

f'(x) = `(sec(5x))'*ln(cos(5x)) + sec(5x)*(ln(cos(5x)))'`

=> `5*sec(5x)*tan x*ln(cos(5x)) + sec(5x)*(-5*(sin 5x)/(cos 5x))`

=> `5*(1/cos(5x))*(sin 5x)/(cos 5x)*ln(cos(5x)) + 1/cos(5x)*(-5*(sin 5x)/(cos 5x))`

=> `5*sin 5x*ln(cos(5x))/(cos^2(5x)) - 5*(sin 5x)/(cos^2 5x)`

**The required derivative is:** `((5*sin 5x)/(cos^2 5x))(ln(cos 5x) - 1)`

never mind i see it now yes it is correct!

can you explain where this line came from as i understand up to the one above it

so i got a different solution to you as i did not get this bit