Question

Can you help me find y'?

What is y' when y = ln(`e^x` + `xe^x`)?

Answer 1

`y' = (2e^ x + xe^x)/(e^x + xe^x) = (x + 2)/(x + 1)`

Explanation:

I would use the chain rule.

`y = ln(e^x+ xe^x)`

If we let `u = e^x+ xe^x`, then by the product rule, we get:

`u' = e^x + 1(e^x) + x(e^x) = 2e^x + xe^x`

We know the derivative of `y = lnu` will be `1/u`,

`y' = 1/u* (2e^x + xe^x)`

`y' = 1/(e^x + xe^x) * (2e^x +xe^x)`

`y' = (2e^ x + xe^x)/(e^x + xe^x)`

We can simplify by factoring out an `e^x`.

`y' = (e^x(2 + x))/(e^x(1 + x))`

`y' = (x + 2)/(x + 1)`

Hopefully this helps!