Question

How do you differentiate `f(x)=tan(lnx)` using the chain rule?

Answer 1

`f'(x)=sec^2(ln(x))/x`

Explanation:

The chain rule states that

`d/dx(f(g(x))=f'(g(x))*g'(x)`

First, we must know that the derivative of `tan(x)` is `sec^2(x)`. With this knowledge, we can create a version of the chain rule specific to tangent functions:

`d/dx(tan(x))=sec^2(x)`

`=>d/dx(tan(g(x))=sec^2(g(x))*g'(x)`

So, if we are differentiating the function `f(x)=tan(ln(x))`, we see that

`f'(x)=sec^2(ln(x))*d/dx(ln(x))`

We must now know that the derivative of `ln(x)` is `1"/"x`, yielding the simplified derivative:

`f'(x)=sec^2(ln(x))*1/x`

`f'(x)=sec^2(ln(x))/x`