What is the derivative of #cos(x^2)#?

1 Answer
Sep 4, 2016

#- 2x sin(x^(2))#

Explanation:

We have: #cos(x^(2))#

This expression can be differentiated using the "chain rule".

Let #u = x^(2) => (d) / (dx) (u) = 2x# and #v = cos(u) => (d) / (du) (cos(u)) = - sin(u)#:

#=> (d) / (dx) (cos(x^(2)) = 2x cdot - sin(u)#

#=> (d) / (dx) (cos(x^(2)) = - 2x sin(u)#

We can now replace #u# with #x^(2)#:

#=> (d) / (dx) (cos(x^(2)) = - 2x sin(x^(2))#