Find y' for y = cube root of (2x²+x-1) = (2x²+x-1)^(1/3)

Find $y'$ for $y = \sqrt[3]{2x^{2}+x-1} = \bigl(2x^{2}+x-1\bigr)^{1/3}$.

Step 1: Apply the chain rule

With $u = 2x^{2}+x-1$ and $n = \dfrac{1}{3}$, so that $u' = 4x+1$: $y' = \frac{1}{3}u^{-2/3}\cdot u' = \frac{1}{3}\bigl(2x^{2}+x-1\bigr)^{-2/3}(4x+1)$.

Step 2: Rewrite in radical form

Since $u^{-2/3} = \dfrac{1}{\sqrt[3]{u^{2}}}$: $y' = \frac{4x+1}{3\,\sqrt[3]{\bigl(2x^{2}+x-1\bigr)^{2}}}$.

Step 3: Verification

\[x = 2:\quad u = 8+2-1 = 9 \;\Longrightarrow\; y' = \frac{9}{3\sqrt[3]{81}} = \frac{3}{\sqrt[3]{81}} \approx 0.69336\]

\[\frac{y(2+10^{-5})-y(2-10^{-5})}{2\times 10^{-5}} \approx 0.69336 \;\;\checkmark\]

Step 4: Where the formula fails

The cube root itself is defined for every real $x$, but the derivative is not. Factoring $2x^{2}+x-1 = (2x-1)(x+1)$, the radicand vanishes at

\[ x = \tfrac12 \qquad\text{and}\qquad x = -1, \]

and at those two points the denominator of $y'$ is zero: the graph has vertical tangents there, exactly as $\sqrt[3]{x}$ does at the origin (compare Problem 24b).