Question

E^4x-1/e^4x+1 derivative

Answer 1

Answer:

not sure what you're asking, the derivative of y=lne4x−1e4x+1 or if you want the derivative of y=ln(e4x)−1e4x+1

Anyway, I'll find both derivative.

This expression y(x)=lne4x−1e4x+1 can be rewritten as :

y(x)=4x−1e4x+1 , because ln(ex)=x

The derivative of 4x−1e4x+1 is pretty easy, it is the derivative of a function over another function :

(uv)′=u′v−uv′v2 , with u and v being functions.

Here u(x)=4x−1;u′(x)=4 and v(x)=e4x+1;v′(x)=4e4x+1 , so you end up with the derivative of y :

y′(x)=4e4x+1−4x+1×4e4x+1(e4x+1)2

⟹y′(x)=−16e4x+1e8x+2

Using the same method for y(x)=ln(e4x)−1e4x+1 , you end up with :

y′(x)=e4x(8−16x)+4e8x+2e4x+1

Hope it was helpful and that I didn't do any mistake.