# How do you find dy/dx by implicit differentiation given e^x=lny?

Jan 2, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{{e}^{x} + x}$

#### Explanation:

Use the following differentiation rules:

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

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

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When we use implicit differentiation, we must differentiate with respect to one variable without isolating another.

The following step represents that we take the derivative with respect to $x$ on each side of the relation.

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

Use the identities above, and implicit differentiation:

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

${e}^{x} / \left(\frac{1}{y}\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = y {e}^{x}$

The relation ${e}^{x} = \ln y$ is simple enough such that we can find an explicit equation for $y$ to resubstitute into the derivative.

${e}^{x} = \ln y$

${e}^{{e}^{x}} = y$

We could have also differentiated the above equation to find $\frac{\mathrm{dy}}{\mathrm{dx}}$, but it's simpler just using implicit differentiation. Since $y = {e}^{{e}^{x}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{{e}^{x}} {e}^{x}$

Use the rule $\left({a}^{n}\right) \left({a}^{m}\right) = {a}^{n + m}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{{e}^{x} + x}$

Hopefully this helps!