Chain rule with a fractional power

Let $f(x) = (x^3 + 2x)^{1/3}$ . Find $f'(1)$ .

Hint

Treat $f(x)$ as an outer power function applied to an inner function: $f(x) = u^{1/3}$ where $u = x^3 + 2x$ . Differentiate the outer function, keeping the inner function unchanged inside, then multiply by the derivative of the inner function.

Step-by-step solution

Write $f(x) = u^{1/3}$ with inner function $u = x^3 + 2x$ . By the chain rule, $f'(x) = \frac{1}{3}u^{-2/3} \cdot u'$ , and since $u' = 3x^2 + 2$ , we get $f'(x) = \frac{1}{3}(x^3+2x)^{-2/3}(3x^2+2).$

Now evaluate at $x=1$ : the inner value is $u(1) = 1^3 + 2(1) = 3$ , and $u'(1) = 3(1)^2+2 = 5$ . Substituting gives $f'(1) = \frac{1}{3}(3)^{-2/3}(5) = \frac{5}{3\cdot 3^{2/3}} = \frac{5}{3^{5/3}}.$

Chain rule with a fractional power

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