# How do you find (d^2y)/(dx^2) for x^3+4y^2=1?

##### 1 Answer
Aug 19, 2017

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{12 x {y}^{2} + 9 {x}^{4}}{16 {y}^{3}}$

#### Explanation:

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$ first. Remember to use the chain rule when differentiating a function of $y$.

$\frac{d}{\mathrm{dx}} \left({x}^{3} + 4 {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(1\right)$

$3 {x}^{2} + 8 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 {x}^{2}}{8 y}$

Now we need to differentiate again. In my opinion, the differentiation is simpler as such:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3}{8} {x}^{2} {y}^{-} 1$

Now we need the product rule:

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{3}{8} \left(\frac{d}{\mathrm{dx}} {x}^{2}\right) {y}^{-} 1 - \frac{3}{8} {x}^{2} \left(\frac{d}{\mathrm{dx}} {y}^{-} 1\right)$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{3}{8} \left(2 x\right) {y}^{-} 1 - \frac{3}{2} {x}^{2} \left(- {y}^{-} 2\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

Recall that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 {x}^{2}}{8 y}$:

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{- 3 x}{4 y} + \frac{3 {x}^{2}}{2 {y}^{2}} \left(\frac{- 3 {x}^{2}}{8 y}\right)$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{- 3 x}{4 y} - \frac{9 {x}^{4}}{16 {y}^{3}}$

Getting a common denominator:

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{12 x {y}^{2} + 9 {x}^{4}}{16 {y}^{3}}$