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

Oct 7, 2016

#### Explanation:

$y = {e}^{u}$

Use part of the chain rule...

$\therefore \frac{\mathrm{dy}}{\mathrm{du}} = {e}^{u} = {e}^{\sqrt{\frac{1}{x}}}$

$u = \sqrt{\frac{1}{x}}$

$u = {\left(\frac{1}{x}\right)}^{\frac{1}{2}}$

${u}^{2} = \frac{1}{x}$

$\ln \left({u}^{2}\right) = \ln \left(\frac{1}{x}\right)$

$2 \ln \left(u\right) = \ln 1 - \ln x$

$2 \ln u = - \ln x$

Now use implicit differentiation...

$\frac{2}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = - \frac{1}{x}$

$\frac{u}{2} \cdot \frac{2}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = - \frac{1}{x} \cdot \frac{u}{2}$

$\frac{\mathrm{du}}{\mathrm{dx}} = - \frac{1}{2 x} \cdot \sqrt{\frac{1}{x}}$

Now use the chain rule...

$\frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = - \frac{1}{2 x} \cdot {e}^{\sqrt{\frac{1}{x}}} \cdot \sqrt{\frac{1}{x}} = \frac{\mathrm{dy}}{\mathrm{dx}}$

Now, you can simplify the final result...

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

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

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

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

Still looks quite ugly, but this is the result you're looking for. It is what it is.

Oct 7, 2016

e^(sqrt(1/x))/(2x^(3/2)

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{chain rule}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots . \left(A\right)$

let $u = \sqrt{\frac{1}{x}} = {\left(\frac{1}{x}\right)}^{\frac{1}{2}} = \frac{1}{x} ^ \left(\frac{1}{2}\right) = {x}^{- \frac{1}{2}}$

differentiate using the $\textcolor{b l u e}{\text{power rule}}$

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = - \frac{1}{2} {x}^{- \frac{3}{2}}$

Now $y = f \left(x\right) = {e}^{u} \Rightarrow \frac{\mathrm{dy}}{\mathrm{du}} = {e}^{u}$

substitute results for $\frac{\mathrm{dy}}{\mathrm{du}} \text{ and } \frac{\mathrm{du}}{\mathrm{dx}}$ into (A) changing u back to x.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{u} . - \frac{1}{2} {x}^{- \frac{3}{2}} = - \frac{1}{2} {x}^{- \frac{3}{2}} {e}^{\sqrt{\frac{1}{x}}}$

rArrdy/dx=(e^(sqrt(1/x)))/(2x^(3/2)