Thread: Trig differentation

How do you derive this equation:

y = sin ^ 3 (cosx^3)

This is the chain rule if the cosine is inside the sine, which what this seems to be.

Originally Posted by mathamatics112

How do you derive this equation:

y = sin ^ 3 (cosx^3)

You need to use the chain rule several times:







And so on. Can you continue?

y= sin^3( cos x^3)
= 3 sin ^ 2 ( cos x^3)(coz(cosx^3))(-sinx^3)(3x^2)
= 9 x^2(sin^2)(cos(cos x^3))(-sin x^3))


are my steps right?

Originally Posted by mathamatics112

y= sin^3( cos x^3)
= 3 sin ^ 2 ( cos x^3)(coz(cosx^3))(-sinx^3)(3x^2)
= 9 x^2(sin^2)(cos(cos x^3))(-sin x^3))

Aren't you missing something in that last line?

i'm not seeing it

Originally Posted by mathamatics112

i'm not seeing it

Did you look at the part that I highlighted? You have sitting by itself. You are taking the sine-squared of what?