# Given y= lnx/e^x how do you find find f'(1)?

Oct 30, 2016

The answer is $= \frac{1}{e}$

#### Explanation:

The function is a quotient of 2 derivable functions $\frac{u}{v}$
And the derivative is $\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{v} ^ 2$

here $u = \ln x$$\implies$$u ' = \frac{1}{x}$
and $v = {e}^{x}$$\implies$$v ' = {e}^{x}$

So $f ' \left(x\right) = \frac{\frac{1}{x} \cdot {e}^{x} - {e}^{x} \ln x}{{e}^{x}} ^ 2 = \frac{{e}^{x} \left(\frac{1}{x} - \ln x\right)}{{e}^{x}} ^ 2$
$= \frac{\frac{1}{x} - \ln x}{e} ^ x$
so $f ' \left(1\right) = \frac{1 - 0}{e} = \frac{1}{e}$