Question

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

Answer 1

`[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)`