Question

Take the derivative of f(x)=(cos(4x))/(1-sin(4x))

Answer 1

You can use the quotient rule:

\(f(x) = \dfrac{g(x)}{h(x) }\)

\(f'(x) = \dfrac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \)

Answer 2

\(g(x) = \cos 4x \) ---> \(g'(x) = -4 \sin 4x\)

\(h(x) = 1 - \sin 4x\) ---> \(h'(x) = -4 \cos 4x\)

Answer 3

Now just substitute and simplify the algebra.

Answer 4

I got this so far:
\[\frac{ -4[\sin4x-\sin^2x-\cos^24x] }{ (1-\sin4x)^2 }\]
@mathstudent55

Answer 5

Looks good, but don't factor a -4 out like that, you're causing more trouble for yourself I think.
Just pull a 4 out.

Answer 6

\[\Large\rm =\frac{4(-\sin4x+\color{orangered}{\sin^24x+\cos^24x})}{(1-\sin4x)^2}\]Notice anything going on here? :)

Answer 7

the red part is equal to 1!

Answer 8

Thank you so much!