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

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Answer 1 · 2016-09-04T08:25:53

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

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))#