How do you differentiate #f(x)=cos(e^(3x^3-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))#

Related questions
See all questions in Chain Rule
Impact of this question
1389 views around the world
You can reuse this answer
Creative Commons License