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