1. ## differentiation again

given $x^{\frac 2 3} + y^{\frac 2 3} = a ^{\frac 2 3}$
find d^2 y / dx^2

my working:
dy/dx = $-(\frac y x )^{1/3}$
d^2 y / dx^2 = ${- \frac 1 3} (\frac y x)^ {- \frac 2 3}$ $\mbox{ }$ ${\frac{x \frac {dy}{dx} - y}{x^2}}$
$= \frac {a^{\frac 2 3}}{3 x^{\frac 4 3} y ^{\frac 1 3}}$
<<<--------------- it's such a lengthy process! i cannot get the answer. is my method completely wrong?

2. It may be advantageous to resist algebraic manipulation before calculating the 2nd derivative. This allows the simpler product rule, rather than the cumbersome quotient rule.

3. Hello, afeasfaerw23231233!

Your work (so far) is correct . . .

Given: . $x^{\frac 2 3} + y^{\frac 2 3} \:= \:a ^{\frac 2 3}$ . . Find $\frac{d^2y}{dx^2 }$

My working: . $\frac{dy}{dx}\:=\:-\left(\frac{y}{x} \right)^{\frac{1}{3}}\;\;{\color{blue}\hdots \text{Yes!}}$

$\frac{d^2y}{dx^2} \:=\: -\frac{1}{3}\left(\frac{y}{x}\right)^{-\frac{2}{3}}{\frac{x \frac{dy}{dx} - y}{x^2}}\;\;{\color{blue}\hdots\text{Right!}}$

Answer: . $\frac {a^{\frac 2 3}}{3 x^{\frac 4 3} y ^{\frac 1 3}}$
Keep going . . .

$\text{We have: }\;\frac{d^2y}{dx^2} \;=\;-\frac{1}{3}\cdot\frac{x^{\frac{2}{3}}}{y^{\frac{2} {3}}} \cdot\frac{x\left(-\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}\right) - y}{x^2}$

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

. . . . . . . . . $= \;\frac{1}{3}\cdot\frac{\overbrace{x^{\frac{2}{3}} + y^{\frac{2}{3}}}^{\text{This is }a^{\frac{2}{3}}}}{x^{\frac{4}{3}}y^{\frac{1}{3}}}$

Therefore: . $\frac{d^2y}{dx^2} \;=\;\frac{a^{\frac{2}{3}}}{3x^{\frac{4}{3}}y^{\fr ac{1}{3}}}$