# How do you find the derivative of y=lne^x?

Dec 5, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1$

#### Explanation:

We should know for this approach that $\frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{1}{x}$.

Applying the chain rule to this derivative tells us that if we were to have a function $u$ instead of just the variable $x$ within the logarithm, we see that $\frac{d}{\mathrm{dx}} \ln \left(u\right) = \frac{1}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

So we see that $\frac{d}{\mathrm{dx}} \ln \left({e}^{x}\right) = \frac{1}{e} ^ x \cdot \frac{d}{\mathrm{dx}} \left({e}^{x}\right)$

Since $\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$:

$\frac{d}{\mathrm{dx}} \ln \left({e}^{x}\right) = \frac{1}{e} ^ x \cdot {e}^{x} = 1$

Dec 5, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1$

#### Explanation:

The logarithm function and exponential functions are inverse functions--they undo one another! This means that ${\log}_{a} \left({a}^{x}\right) = x$ and ${a}^{{\log}_{a} \left(x\right)} = x$.

Recall that the function $\ln \left(x\right)$ is the logarithm with a base of $e$, that is, $\ln \left(x\right) = {\log}_{e} \left(x\right)$. Thus:

$y = \ln \left({e}^{x}\right) = {\log}_{e} \left({e}^{x}\right) = x$

So

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1$