# How do you use implicit differentiation to find dy/dx given ysqrt(x+1)=4?

Nov 19, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 4 \sqrt{x + 1}}{x + 1} ^ 2$

#### Explanation:

$y \sqrt{x + 1} = 4$

Applying implicit differentiation:

$y \cdot {\left(x + 1\right)}^{- \frac{1}{2}} \cdot 1 + {\left(x + 1\right)}^{\frac{1}{2}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$ (Product Rule)

From the original equation: $y = \frac{4}{\sqrt{x + 1}}$

Hence: ${\left(x + 1\right)}^{\frac{1}{2}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 4}{\sqrt{x + 1}} \cdot {\left(x + 1\right)}^{- \frac{1}{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 4}{x + 1} \cdot \frac{1}{\sqrt{x + 1}}$

$= \frac{- 4 \sqrt{x + 1}}{x + 1} ^ 2$