# Question ce2e0

Nov 12, 2016

Although we could solve for $y$, this looks like a classic implicit differentiation exercise.

#### Explanation:

I assume that you want the second derivative of $y$ with respect to $x$. (Yes, there are other possibilities.)

${x}^{2} + {y}^{2} = 90$

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{d}{\mathrm{dx}} \left({x}^{2} + {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(90\right)$

$2 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$, so

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y}$

Differentiate again:

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

Simplify a bit,

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

Replace $\frac{\mathrm{dy}}{\mathrm{dx}}$ with its equivalent $- \frac{x}{y}$

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

Simplify again

$= \frac{- y - {x}^{2} / y}{y} ^ 2$

Clear the fraction in the nmumerator

$= \frac{\left(- y - {x}^{2} / y\right) \cdot y}{{y}^{2} \cdot y}$

$= \frac{- {y}^{2} - {x}^{2}}{y} ^ 3$

Factor out a -1)

$= \frac{- \left({y}^{2} + {x}^{2}\right)}{y} ^ 3$

Use the original rfelationship ${x}^{2} + {y}^{2} = 90$ to finish simplifying.

$= \frac{- 90}{y} ^ 3$