# How do you find dy/dx given x=1/(1+sqrty)?

Feb 25, 2017

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

#### Explanation:

differentiate both sides $\textcolor{b l u e}{\text{implicitly with respect to x}}$

$\text{Express "1/(1+y^(1/2))=(1+y^(1/2))^-1" and}$

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

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

$\Rightarrow 1 = - {\left(1 + {y}^{\frac{1}{2}}\right)}^{-} 2 \times \frac{1}{2} {y}^{- \frac{1}{2}} . \frac{\mathrm{dy}}{\mathrm{dx}}$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2 \sqrt{y} {\left(1 + \sqrt{y}\right)}^{2}}$