# How do you find (d^2y)/(dx^2) for 5x+3y^2=1?

Jun 28, 2017

#### Explanation:

Using the sum rule and chain rule to differentiate both sides with respect to $x$ gives:

$5 + 6 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Differentiating again using the product rule (chain rule is applied in the first term, hence the squared derivative):

$6 \cdot {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} + 6 y \cdot \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$

Rearranging the equation for the first derivative:

$6 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = - 5$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{5}{6 y}$

Substituting this into the equation for the second derivative:

$\frac{150}{36 {y}^{2}} + 6 y \cdot \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$

$6 y \cdot \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{150}{36 {y}^{2}}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{25}{36 {y}^{3}}$

There you go. If you want it to be in terms of $x$, then you'll end up with some $\pm$'s in there. Of course, when $y = 0$, the derivative is undefined, which makes sense, since the tangent line at that point is vertical, with infinite slope.