# How do you differentiate f(x)=e^(4^(1/(x^2))) using the chain rule?

Nov 11, 2015

I read this as $e$ to the power of ($4$ raised to the $\frac{1}{x} ^ 2$ power or ${4}^{{x}^{-} 2}$). For the derivative of $f \left(x\right) = {e}^{{4}^{\frac{1}{x} ^ 2}} = {e}^{{4}^{{x}^{-} 2}}$, see below.

#### Explanation:

The derivative of ${e}^{u}$ with respect to $x$ is ${e}^{u} \frac{\mathrm{du}}{\mathrm{dx}}$.

So the first step gives us:

$f ' \left(x\right) = {e}^{{4}^{{x}^{-} 2}} \frac{d}{\mathrm{dx}} \left({4}^{{x}^{-} 2}\right)$

The derivative with respect to $x$ of ${4}^{u}$ is ${4}^{u} \ln 4 \frac{\mathrm{du}}{\mathrm{dx}}$

So our second step gets us:

$f ' \left(x\right) = {e}^{{4}^{{x}^{-} 2}} \left({4}^{{x}^{-} 2} \ln 4 \frac{d}{\mathrm{dx}} \left({x}^{-} 2\right)\right)$

And the third step gets us:

$f ' \left(x\right) = {e}^{{4}^{{x}^{-} 2}} \left({4}^{{x}^{-} 2} \ln 4 \left(- 2 {x}^{-} 3\right)\right)$

We can clean this up a little bit:

$f ' \left(x\right) = - \frac{2}{x} ^ 3 {4}^{{x}^{-} 2} {e}^{{4}^{{x}^{-} 2}} \ln 4$

$= - \frac{2}{x} ^ 3 {4}^{\frac{1}{x} ^ 2} {e}^{{4}^{\frac{1}{x} ^ 2}} \ln 4$

Other expressions are also possible.