Question

if (f(x)=(x^{2}-2x - 1)^{\frac{1}{3}}), then (f(0)) is
(a) (\frac{4}{3})
(b) 0
(c) (-\frac{2}{3})
(d) (\frac{1}{3})
(e) 2

Explanation:

Step1: Apply chain - rule

Let $u = x^{2}-2x - 1$, then $y = u^{\frac{1}{3}}$. The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$ and $\frac{du}{dx}$.
$\frac{dy}{du}=\frac{1}{3}u^{-\frac{2}{3}}$ and $\frac{du}{dx}=2x - 2$.
So, $f^{\prime}(x)=\frac{1}{3}(x^{2}-2x - 1)^{-\frac{2}{3}}\cdot(2x - 2)$.

Step2: Evaluate at $x = 0$

Substitute $x = 0$ into $f^{\prime}(x)$.
$f^{\prime}(0)=\frac{1}{3}(0^{2}-2\times0 - 1)^{-\frac{2}{3}}\cdot(2\times0 - 2)$.
Since $(0^{2}-2\times0 - 1)^{-\frac{2}{3}}=(-1)^{-\frac{2}{3}} = 1$, then $f^{\prime}(0)=\frac{1}{3}\times1\times(-2)=-\frac{2}{3}$.

Answer:

C. $-\frac{2}{3}$