Find y' for x^(2/3)+y^(2/3)=a^(2/3) — the astroid

Find $y'$ for $x^{2/3}+y^{2/3}=a^{2/3}$ — the astroid.

Step 1: Differentiate both sides with respect to $x$

Each term is a power, and the $y$ term picks up a factor $y'$: $\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\,y' = 0$.

The right-hand side is a constant, so its derivative vanishes.

Step 2: Solve for $y'$

The factors $\tfrac23$ cancel, and dividing negative powers inverts the ratio: \[y' = -\frac{x^{-1/3}}{y^{-1/3}} = -\left(\frac{y}{x}\right)^{1/3} = -\sqrt[3]{\frac{y}{x}}.\]

Step 3: Verification against the parametric form

Problem 195 parametrised the astroid as $x = a\cos^{3}t$, $y = a\sin^{3}t$ and found the slope $-\tan t$. The two answers agree: \[-\sqrt[3]{\frac{y}{x}} = -\sqrt[3]{\frac{a\sin^{3}t}{a\cos^{3}t}} = -\frac{\sin t}{\cos t} = -\tan t \;\;\checkmark\]

\[a = 1,\ x = y = \tfrac{1}{2\sqrt2}:\quad y' = -\sqrt[3]{1} = -1, \text{ the symmetry point on } y = x \;\;\checkmark\]

Step 4: The four cusps

At $(\pm a,0)$ the slope is $0$ and at $(0,\pm a)$ the formula breaks down with a vertical tangent. These four points are the cusps of the astroid, where the curve has a genuine corner rather than a smooth turning point.