How do you differentiate given #tan^2(x)#?

1 Answer
Nov 23, 2016

Answer:

#d/dx[tan^2x]=2tanxsec^2x#.

Explanation:

The big idea here is that we are squaring something. The derivative of #x^2# is #2x#, so the derivative of #u# squared (#u# stands for any function) is #2u# * (the derivative of #u#).

This is by the chain rule.

In other words, #d/dx[u^2]=2u*d/dx[u]#.

So #d/dx[tan^2x]=2tanx*d/dx[tanx]#.

Remember that #d/dx[tanx]=sec^2x#, so our final answer is #d/dx[tan^2x]=2tanxsec^2x#.

If you don't remember why or how #d/dx[tanx]=sec^2x#, rewrite #tanx=sinx/cosx# and take the derivative using the quotient rule.

#d/dx[tanx]=d/dx[sinx/cosx]=(cos^2x-(-sin^2x))/cos^2x=1/cos^2x=sec^2x#