# How do you determine (d^2y)/(dx^2) given x^(2/3)+y^(2/3)=8?

Jan 6, 2017

Implicit differentiation, stubbornness and attention to details.

#### Explanation:

The more challenging (difficult) a problem is, the MORE you should write. (As a student I tried to skip steps and it took me a year of two to learn this important fact. If you can answer in two lines, do it in one. If you think it will take 8 lines, plan to write 12.)

${x}^{\frac{2}{3}} + {y}^{\frac{2}{3}} = 8$

Differentiate

$\frac{2}{3} {x}^{- \frac{1}{3}} + \frac{2}{3} {y}^{- \frac{1}{3}} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Multiply through b $\frac{3}{2}$ to get rid of those fractions.

${x}^{- \frac{1}{3}} + {y}^{- \frac{1}{3}} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$.

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

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

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

It's usually silly to have a negative exponent in a fraction, so let's fix that.

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

Now differentiate again.

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

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

Let's factor out that $\frac{1}{3}$ (We could take a $- \frac{1}{3}$, but $\frac{\mathrm{dy}}{\mathrm{dx}}$ has a minus in front of it.)

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

Use $\frac{\mathrm{dy}}{\mathrm{dx}} = - {y}^{\frac{1}{3}} / {x}^{\frac{1}{3}}$ to get

$= \frac{1}{3} \frac{- \left({x}^{\frac{1}{3}} / {y}^{\frac{2}{3}}\right) \left(- {y}^{\frac{1}{3}} / {x}^{\frac{1}{3}}\right) + \left({y}^{\frac{1}{3}} / {x}^{\frac{2}{3}}\right)}{x} ^ \left(\frac{2}{3}\right)$

$= \frac{1}{3} \frac{\left(\frac{1}{y} ^ \left(\frac{1}{3}\right)\right) + \left({y}^{\frac{1}{3}} / {x}^{\frac{2}{3}}\right)}{x} ^ \left(\frac{2}{3}\right)$

Clear the denominators in the numerator by multiplying the big fraction by $\frac{\left({x}^{\frac{2}{3}} {y}^{\frac{1}{3}}\right)}{\left({x}^{\frac{2}{3}} {y}^{\frac{1}{3}}\right)}$

$= \frac{1}{3} \frac{\left(\left(\frac{1}{y} ^ \left(\frac{1}{3}\right)\right) + \left({y}^{\frac{1}{3}} / {x}^{\frac{2}{3}}\right)\right)}{{x}^{\frac{2}{3}}} \cdot \frac{\left({x}^{\frac{2}{3}} {y}^{\frac{1}{3}}\right)}{\left({x}^{\frac{2}{3}} {y}^{\frac{1}{3}}\right)}$

$= \frac{1}{3} \frac{{x}^{\frac{2}{3}} + {y}^{\frac{2}{3}}}{{x}^{\frac{4}{3}} {y}^{\frac{1}{3}}}$

Finally, notice that the numerator is the expression we started with and is equal to $8$.

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{8}{3 {x}^{\frac{4}{3}} {y}^{\frac{1}{3}}}$

Jan 6, 2017

$y ' ' = \frac{1}{3} {y}^{\frac{1}{3}} {x}^{- \frac{4}{3}} - \frac{1}{3} {x}^{- \frac{1}{3}} {y}^{- \frac{2}{3}} y '$

#### Explanation:

Differentiate the equation with respect to $x$:

$\frac{d}{\mathrm{dx}} \left({x}^{\frac{2}{3}} + {y}^{\frac{2}{3}}\right) = 0$

$\frac{2}{3} {x}^{- \frac{1}{3}} + \frac{2}{3} {y}^{- \frac{1}{3}} y ' = 0$

Solving for $y '$ we have:

$y ' = - {x}^{- \frac{1}{3}} / {y}^{- \frac{1}{3}} = - {y}^{\frac{1}{3}} / {x}^{\frac{1}{3}}$

Differentiating again:

$y ' ' = - \frac{\frac{1}{3} {x}^{\frac{1}{3}} {y}^{- \frac{2}{3}} y ' - \frac{1}{3} {y}^{\frac{1}{3}} {x}^{- \frac{2}{3}}}{x} ^ \left(\frac{2}{3}\right) = \frac{1}{3} {y}^{\frac{1}{3}} {x}^{- \frac{4}{3}} - \frac{1}{3} {x}^{- \frac{1}{3}} {y}^{- \frac{2}{3}} y '$