Question

How do you differentiate `f(x)=(lnx+4x)(tanx-e^x)` using the product rule?

Answer 1

`f'(x) = (lnx+4x)(sec^2x-e^x) + (tanx-e^x)(1/x+4)`

Explanation:

You need to use the product rule; `d/dx(uv)=u(dv)/dx+v(du)/dx`

We have `f(x)=(lnx+4x)(tanx-e^x)` with `u=(lnx+4x)`; `v=(tanx-e^x)`

So, `f'(x) = (lnx+4x)d/dx(tanx-e^x) + (tanx-e^x)d/dx(lnx+4x)`

`:. f'(x) = (lnx+4x)(sec^2x-e^x) + (tanx-e^x)(1/x+4)`