How do you take the derivative of #Tan(2x+1)#?

1 Answer

#[tan(2x+1)]^'=2/(cos^2(2x+1))=2sec^2(2x+1)#

Explanation:

We can use the chain rule : if #f# and #g# are differentiable functions, then the rule states that
#[f[g(x)]]^'=f^'[g(x)]g^'(x)#

In our specific case:
#f(u)=tan(u)#
#g(x)=2x+1#
The derivatives of these two functions are:
#f^'(u)=1/(cos^2(u))=sec^2(u)#
#g^'(x)=2#

In the end, you get that
#[tan(2x+1)]^'=[f[g(x)]]^'=f^'[g(x)]g^'(x)=1/(cos^2(2x+1))*2=2/(cos^2(2x+1))=2sec^2(2x+1)#