Calculus help?

[namarino]

Would someone be willing to help me with this calculus problem? I have to take the derivative of the following:

 

j(x) = (cos(x-x^3))^3

 

And if you would't mind showing your steps, I'd really appreciate it. Thanks so much!!

[Enderman]

this shows you all the steps:

http://www.derivative-calculator.net/

[T.Vengeance]

Chain rule it. So

 

Derivative of the outer (power rule) with the inner, times the derivative of your inner.

 

So the derivative of u^3 = 3u^2. Substitute cosine(x-x^3) back into u.

 

The inner function isn't a pure cosine function either so you have to chain rule that as well.

 

Derivative of cos(x-x^3) = -sin(x-x^3)*(1-3x^2)

 

Once again derivative of the outer function aka cosine with the inner function, the x-x^3. Times the derivative of the inner function (x-x^3) which is 1-3x^2.

 

3*cos(x-x^3)^2 * (-sin(x-x^3))*(1-3x^2)