# How do you use the chain rule to differentiate y=sqrt(1/(x+1))?

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2 {\left(x + 1\right)}^{\frac{3}{2}}}$

#### Explanation:

Given function:

$y = \setminus \sqrt{\frac{1}{x + 1}}$

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

$y = {\left(x + 1\right)}^{- \frac{1}{2}}$

Differentiating above function w.r.t. $x$ using chain rule as follows

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({\left(x + 1\right)}^{- \frac{1}{2}}\right)$

$= - \frac{1}{2} {\left(x + 1\right)}^{- \frac{1}{2} - 1} \frac{d}{\mathrm{dx}} \left(x + 1\right)$

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

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

Jul 22, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2 \left(x + 1\right) \sqrt{x + 1}} = - \frac{1}{2 {\left(x + 1\right)}^{\frac{3}{2}}} = - \frac{1}{2 {\left(\sqrt{x + 1}\right)}^{3}}$.

#### Explanation:

$y = \sqrt{\frac{1}{x + 1}} = \frac{1}{\sqrt{x + 1}}$.

Let, $\left(x + 1\right) = u , \sqrt{u} = \sqrt{x + 1} = v$.

Thus, $y = \frac{1}{v} , v = \sqrt{u} , u = x + 1$.

Thus, $y$ is a function of $v , v \text{ of "u, &, u" of } x$.

Therefore, by the Chain Rule,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}} \ldots \ldots \ldots . \left(\ast\right)$.

Now, $\frac{\mathrm{dy}}{\mathrm{dv}} = \frac{d}{\mathrm{dv}} \left\{\frac{1}{v}\right\} = - \frac{1}{v} ^ 2. \ldots . . \left({\ast}^{1}\right)$,

$\frac{\mathrm{dv}}{\mathrm{du}} = \frac{d}{\mathrm{du}} \left\{\sqrt{u}\right\} = \frac{1}{2 \sqrt{u}} \ldots \ldots . . \left({\ast}^{2}\right)$, and,

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left\{x + 1\right\} = 1. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left({\ast}^{3}\right)$.

We combine $\left({\ast}^{1}\right) , \left({\ast}^{2}\right) , \left({\ast}^{3}\right) \mathmr{and} \left(\ast\right)$, to get,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(- \frac{1}{v} ^ 2\right) \left(\frac{1}{2 \sqrt{u}}\right) \left(1\right) ,$

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2 \left(x + 1\right) \sqrt{x + 1}} = - \frac{1}{2 {\left(x + 1\right)}^{\frac{3}{2}}} = - \frac{1}{2 {\left(\sqrt{x + 1}\right)}^{3}}$.