How do you differentiate f(x)=cos(e^(3x^3-x)) f(x)=cos(e3x3−x) using the chain rule?

1 Answer
sente
Jun 21, 2016

f'(x) = -(9x^2-1)e^(3x^3-x)sin(e^(3x^3-x))

Explanation:

Using the sum rule and the chain rule, along with the following derivatives:

  • d/dxcos(x) = -sin(x)

  • d/dxe^x = e^x

  • d/dx x^n = nx^(n-1)

we have

f'(x) = d/dxcos(e^(3x^3-x))

=-sin(e^(3x^3-x))(d/dxe^(3x^3-x))

=-sin(e^(3x^3-x))e^(3x^3-x)(d/dx(3x^3-x))

=-sin(e^(3x^3-x))e^(3x^3-x)((d/dx3x^3)-(d/dxx))

=-sin(e^(3x^3-x))e^(3x^3-x)(9x^2-1)

:. f'(x) = -(9x^2-1)e^(3x^3-x)sin(e^(3x^3-x))

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