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

Jan 17, 2016

Use $\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \frac{\mathrm{du}}{\mathrm{dx}}$ and $\frac{d}{\mathrm{dx}} \left(\sqrt{u}\right) = \frac{1}{2 \sqrt{u}} \frac{\mathrm{du}}{\mathrm{dx}}$ or rewrite the function first.

Explanation:

Using the chain rule to differentiate the function as presented:

$f ' \left(x\right) = {e}^{\sqrt{\frac{1}{x} ^ 2}} \frac{d}{\mathrm{dx}} \left(\sqrt{\frac{1}{x} ^ 2}\right)$

$= {e}^{\sqrt{\frac{1}{x} ^ 2}} \cdot \frac{1}{2 \sqrt{\frac{1}{x} ^ 2}} \frac{d}{\mathrm{dx}} \left(\frac{1}{x} ^ 2\right)$

$= {e}^{\sqrt{\frac{1}{x} ^ 2}} \cdot \frac{1}{2 \sqrt{\frac{1}{x} ^ 2}} \left(\frac{- 2}{x} ^ 3\right)$

$= \frac{- \sqrt{{x}^{2}} \cdot {e}^{\sqrt{\frac{1}{x} ^ 2}}}{x} ^ 3$

Rewriting the function as a piecewise function first

Since ${\sqrt{x}}^{2} = \left\mid x \right\mid = \left\{\begin{matrix}- x & \text{ & " & x < 0 \\ x & " & } & 0 < x\end{matrix}\right.$, we can rewrite the function as:

$f \left(x\right) = \left\{\begin{matrix}{e}^{- \frac{1}{x}} & \text{ & " & x < 0 \\ e^(1/x) & " & } & 0 < x\end{matrix}\right.$

Differentiating each part gets us:

$f ' \left(x\right) = \left\{\begin{matrix}{e}^{\frac{1}{x}} / {x}^{2} & \text{ & " & x < 0 \\ -e^(1/x)/x^2 & " & } & 0 < x\end{matrix}\right.$

This is equivalent to the answer given above.