# What is the derivative of e^(lnx)?

May 29, 2017

$1$

#### Explanation:

We can also do this without first using the identity ${e}^{\ln} x = x$, although we will have to use this eventually.

Note that $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$, so when we have a function in the exponent the chain rule will apply: $\frac{d}{\mathrm{dx}} {e}^{u} = {e}^{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

So:

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

The derivative of $\ln x$ is $\frac{1}{x}$:

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

Then using the identity ${e}^{\ln} x = x$:

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

Which is the same as the answer we'd get if we use the identity from the outset (which is what I recommend you do--this is just a fun way to show that "calculus works".)

$\frac{d}{\mathrm{dx}} {e}^{\ln} x = \frac{d}{\mathrm{dx}} x = 1$