Question

if f(x)=1-xe^x/x+e^x,find the derivatives of f(x)​

Answer 1

$\color{red}\huge{\underline{\underline{\bf GIVEN\dag}}}$

$f(x) = \frac{1 - {e}^{x} }{x + {e}^{x} }$

$\color{red}\huge{\underline{\underline{\bf SOLUTION \dag}}}$

This is a problem that is using the quotient rule

the formula for this that you have to remember is:

$\blue{ \boxed{ \frac{u ^ ′v−uv^′}{v^{2}}}}$

The first things that you have to do is determining the derivatives of each of the functions that you have.

$u=1−xe^ x→u^ ′=e^ x+xe ^ x$

 (remembering that you have to do the product rule in the process u′v+uv′ and that the derivative of a constant = 0)

$v=x+ex→v ^′=1+e ^x$

Then you can now just plug all of your u and v and their derivatives in their proper spot and you got your answer:

$\orange{ \therefore f^ ′(x) = \frac{(e {}^x+xe {}^x)(x+e ^ x)−(1−xe ^ x)(1+e {}^ x)}{(x+e ^ x)^2}}$

Usually on exams or tests, they will not ask you not to simplify but if you do need to simplify, factor the $e^x$ on the top side. And remember when simplifying, never do anything to the bottom.

HOPE THIS HELPS!!