Solve Derivative of sin(cos(cuberoot(x^2-1)))

[menal]

Homework Statement

sin(cos(cuberoot(x^2-1))

i noe you change it to sin(cos(x^2-1)^1/3)
and then (cos(cos(x^2-1)^1/3)(-sin(x^2-1)^1/3)(1/… (2x)
amd then you factor out (x^2-1)^-2/3
and are left with (cos(cos(x^2-1))(-sin(x^2-1))(1/3)(2x)
but what do you do afterwards?

 

[Undoubtedly0]

Greetings! Is the problem asking to take the derivative of sin(cos((x²-1)^1/3)? If so, it looks like you're on the right track, although I'm not sure why (x²-1)^-2/3 was factored out. Just keep applying the chain rule.

\frac{d}{dx}[\sin(\cos((x^2-1)^{1/3}))]

=\cos(\cos((x^2-1)^{1/3})) \frac{d}{dx}[\cos((x^2-1)^{1/3}))]

=\cos(\cos((x^2-1)^{1/3})) (-\sin((x^2-1)^{1/3}))\frac{d}{dx}[(x^2-1)^{1/3}]

=-\cos(\cos((x^2-1)^{1/3}))\sin((x^2-1)^{1/3})((1/3)(x^2-1)^{-2/3})\frac{d}{dx}[x^2-1]


=-\cos(\cos((x^2-1)^{1/3}))\sin((x^2-1)^{1/3})((1/3)(x^2-1)^{-2/3})(2x)


=\frac{-2x\cos(\cos((x^2-1)^{1/3}))\sin((x^2-1)^{1/3})}{3(x^2-1)^{2/3}}