Question

Differentiate.
f(t) = $\frac{\sqrt[3]{x}}{x-6}$

Answer 1

Answer:

The derivative with respect to $x$ is:

$\displaystyle \frac{-2\, x - 6}{3\, x^{2/3}\, (x - 6)^{2}}$;

Or equivalently:

$\displaystyle \frac{-2\, x - 6}{3\, \sqrt[3]{x^{2}}\, (x - 6)^{2}}$.

Step-by-step explanation:

Assume that the question is asking for the derivative of $(\sqrt[3]{x}) / (x - 6)$ with respect to $x$. Apply the following steps:

• Rewrite the cubic root $\sqrt[3]{x}$ as $x$ raised to the fractional power $(1/3)$, so as to fit the form of the power rule.
• Apply the quotient rule to find the derivative. Alternatively, rewrite the denominator $1 / (x - 6)$ as $(x - 6)^{-1}$ and apply the product rule instead.

By the power rule, for all real number $r$, including fractions such as $(1/3)$:

$\displaystyle \frac{d}{d x} \left[x^{r}\right] = r\, x^{r - 1}$.

Hence:

$\displaystyle \frac{d}{d x} \left[\sqrt[3]{x}\right] = \frac{d}{d x} \left[x^{1/3}\right] = \frac{1}{3}\, x^{-2/3}$.

Since $x^{-r} = (1 / x^{r})$ for all non-zero $x$:

$\displaystyle \frac{1}{3}\, x^{-2/3} = \frac{1}{3\, x^{2/3}}$.

The derivative of the original expression would be:

$\begin{aligned} & \frac{d}{d x} \left[\frac{\sqrt[3]{x}}{x - 6}\right] \\ =\; & \frac{d}{d x}\left[\left(x^{1/3}\right)\, (x - 6)^{-1}\right] \\ =\; & \frac{d}{d x}\, \left[\left(x^{1/3}\right)\right]\, (x - 6)^{-1} + \left(x^{1/3}\right)\, \frac{d}{d x}\, \left[(x - 6)^{-1}\right] \\ =\; & \left(\frac{1}{3\, x^{2/3}}\right)\, (x - 6)^{-1} + \left(x^{1/3}\right)\, (-1)\, (x - 6)^{-2} \end{aligned}$.

Make use of the fact that $x^{-r} = (1 / x^{r})$ for all non-zero $x$ to further simplify the expression:

$\begin{aligned} & \left(\frac{1}{3\, x^{2/3}}\right)\, (x - 6)^{-1} + \left(x^{1/3}\right)\, (-1)\, (x - 6)^{-2} \\ =\; &\frac{1}{\left(3\, x^{2/3}\right)\, (x - 6)} + \frac{(-1)\, x^{1/3}}{\, (x - 6)^{2}} \\ =\; & \frac{(x - 6) + ((-1)\, (x^{1/3}))\, (3\, x^{2/3})}{3\, x^{2/3}\, (x - 6)^{2}} \\ =\; & \frac{x - 6 - 3\, x}{3\, x^{2/3}\, (x - 6)^{2}} \\ =\; & \frac{-2\, x - 6}{3\, x^{2/3}\, (x - 6)^{2}} \\ =\; & \frac{-2\, x - 6}{3\, \sqrt[3]{x^{2}}\, (x - 6)^{2}}\end{aligned}$.