Power rule with a fractional exponent

Let $f(x) = x^{2/3}$ . Find $f'(x)$ using the power rule, and then evaluate $f'(8)$ .

Hint

Apply the power rule $\frac{d}{dx}[x^n] = n x^{n-1}$ with $n = \frac{2}{3}$ , then substitute $x = 8$ into the resulting expression. Recall that $8^{1/3} = 2$ .

Step-by-step solution

By the power rule, $\frac{d}{dx}[x^n] = n x^{n-1}$ . Here $n = \frac{2}{3}$ , so $f'(x) = \frac{2}{3} x^{(2/3) - 1} = \frac{2}{3} x^{-1/3}$ .

Now substitute $x = 8$ : $f'(8) = \frac{2}{3} \cdot 8^{-1/3}$ . Since $8^{1/3} = 2$ , we have $8^{-1/3} = \frac{1}{2}$ , so $f'(8) = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}$ .

Power rule with a fractional exponent

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