What is the derivative of #sin x^2#?

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Answer 1 · 2018-04-22T08:54:56

#2xcos(x^2)#

Using the chain rule,

#dy/dx=dy/(du)*(du)/dx#

Let #u=x^2,du=2x \ dx,:.(du)/dx=2x#.

Then #y=sin(u),dy=cos(u) \ du,:.dy/(du)=cos(u)#.

Combining, we get,

#dy/dx=cos(u)*2x#

#=2xcos(u)#

Substitute back #u=x^2# to get the final answer:

#=2xcos(x^2)#