How do you find (d^2y)/(dx^2) for 5x^3=-4y^2+4?

Aug 20, 2016

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{240 x {y}^{2} + 225 {x}^{4}}{64 {y}^{3}}$

Explanation:

First, we will use implicit differentiation to find $\frac{\mathrm{dy}}{\mathrm{dx}}$. Recall that implicit differentiation involves using the chain rule when differentiating a term with $y$.

Differentiating:

$15 {x}^{2} = - 8 y \frac{\mathrm{dy}}{\mathrm{dx}}$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{15 {x}^{2}}{8 y}$

Differentiate again. This time the quotient rule will be used!

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{30 x \left(8 y\right) - 15 {x}^{2} \left(8 \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{8 y} ^ 2$

Replace $\frac{\mathrm{dy}}{\mathrm{dx}}$ with $- \frac{15 {x}^{2}}{8 y}$:

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{30 x \left(8 y\right) - 15 {x}^{2} \left(8 \left(- \frac{15 {x}^{2}}{8 y}\right)\right)}{8 y} ^ 2$

Simplify:

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{240 x y - 15 {x}^{2} \left(- \frac{15 {x}^{2}}{y}\right)}{64 {y}^{2}}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{240 x {y}^{2} - 15 {x}^{2} \left(- 15 {x}^{2}\right)}{64 {y}^{3}}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{240 x {y}^{2} + 225 {x}^{4}}{64 {y}^{3}}$