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

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Answer 1 · 2016-10-03T00:46:44

#y=tan(x^2)=tan(u)#

#:. (dy)/(du)=sec^2(u)=sec^2(x^2)#

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

Use the chain rule...

#(dy)/(du)*(du)/(dx)=2x*sec^2(x^2)=(dy)/(dx)#

Remember that:

If #y=tanx#, #(dy)/(dx)=sec^2(x)#

It's a general rule.