# How do you find (dy)/(dx) given sqrty+xy^2=5?

Mar 15, 2018

$\textcolor{b l u e}{- \frac{2 {y}^{\frac{5}{2}}}{1 + 4 x {y}^{\frac{3}{2}}}}$

#### Explanation:

We need to differentiate this implicitly, because we don't have a function in terms of one variable.

When we differentiate $y$ we use the chain rule:

$\frac{d}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}}$

As an example if we had:

${y}^{2}$

This would be:

$\frac{d}{\mathrm{dy}} \left({y}^{2}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$

In this example we also need to use the product rule on the term $x {y}^{2}$

Writing $\sqrt{y}$ as ${y}^{\frac{1}{2}}$

${y}^{\frac{1}{2}} + x {y}^{2} = 5$

Differentiating:

$\frac{1}{2} {y}^{- \frac{1}{2}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + x \cdot 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + {y}^{2} = 0$

$\frac{1}{2} {y}^{- \frac{1}{2}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + x \cdot 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = - {y}^{2}$

Factor out $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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

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

dy/dx=(-y^2)/((1/2y^(-1/2)+2xy))=(-y^2)/(1/(2sqrt(y))+2xy

Simplify:

Multiply by: $2 \sqrt{y}$

(-y^2*2sqrt(y))/(2sqrt(y)1/(2sqrt(y))+2xy*2sqrt(y)

(-y^2*2sqrt(y))/(cancel(2sqrt(y))1/(cancel(2sqrt(y)))+2xy*2sqrt(y)

$\frac{- {y}^{2} \cdot 2 \sqrt{y}}{1 + 2 x y \cdot 2 \sqrt{y}} = - \frac{2 \sqrt{{y}^{5}}}{1 + 4 x \sqrt{{y}^{3}}} = \textcolor{b l u e}{- \frac{2 {y}^{\frac{5}{2}}}{1 + 4 x {y}^{\frac{3}{2}}}}$