Find y' for y = cube root of (1- cube root of (1

Find $y'$ for $y = \sqrt[3]{1-\sqrt[3]{1-\sqrt{x}}}$.

Step 1: Name the layers

Working from the inside out, set

\[ w = \sqrt{x} = x^{1/2}, \qquad v = 1-w, \qquad u = 1-v^{1/3}, \qquad y = u^{1/3}. \]

So $y$ is a chain of four functions, and the derivative is the product of four factors.

Step 2: Differentiate each layer

\[ \frac{dy}{du} = \frac{1}{3}u^{-2/3}, \qquad \frac{du}{dv} = -\frac{1}{3}v^{-2/3}, \qquad \frac{dv}{dw} = -1, \qquad \frac{dw}{dx} = \frac{1}{2\sqrt{x}}. \]

Step 3: Multiply the factors

By the chain rule $\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dv}\cdot\dfrac{dv}{dw}\cdot\dfrac{dw}{dx}$. The two minus signs cancel: \[y' = \frac{1}{3}u^{-2/3}\cdot\left(-\frac{1}{3}v^{-2/3}\right)\cdot(-1)\cdot\frac{1}{2\sqrt{x}} = \frac{u^{-2/3}\,v^{-2/3}}{18\sqrt{x}}.\]

Step 4: Substitute back

\[ y' = \frac{1}{18\sqrt{x}\,\bigl(1-\sqrt{x}\bigr)^{2/3}\bigl(1-\sqrt[3]{1-\sqrt{x}}\,\bigr)^{2/3}}. \]

Step 5: Verification

\[x = \tfrac{1}{4}:\quad w = \tfrac12,\ v = \tfrac12,\ \sqrt[3]{v} \approx 0.79370,\ u \approx 0.20630\]

\[y' \approx \frac{1}{18(0.5)(0.5)^{2/3}(0.20630)^{2/3}} = \frac{1}{9\cdot 0.62996\cdot 0.34914} \approx 0.50518\]

\[\frac{y(0.25+10^{-6})-y(0.25-10^{-6})}{2\times 10^{-6}} \approx 0.50518 \;\;\checkmark\]

Step 6: Domain of the formula

The nested cube roots are defined for every $x \geq 0$, but the derivative requires all three denominators to be non-zero: $x \neq 0, \qquad 1-\sqrt{x} \neq 0\ \ (x \neq 1), \qquad 1-\sqrt[3]{1-\sqrt{x}} \neq 0\ \ (x \neq 0)$.

So the formula holds on $x \gt 0$ with $x \neq 1$; at $x = 0$ and $x = 1$ the graph has vertical tangents.