QuestionJuly 15, 2025

Find the derivative of the function. f(x)=sqrt [3](x)(x+1)

Find the derivative of the function. f(x)=sqrt [3](x)(x+1)
Find the derivative of the function. f(x)=sqrt [3](x)(x+1)

Solution
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Answer

f'(x) = \frac{1}{3}x^{-2/3}(x+1) + x^{1/3} Explanation 1. Apply the Product Rule The derivative of f(x) = u(x)v(x) is f'(x) = u'(x)v(x) + u(x)v'(x). Here, u(x) = \sqrt[3]{x} and v(x) = x+1. 2. Differentiate u(x) = \sqrt[3]{x} u(x) = x^{1/3}, so u'(x) = \frac{1}{3}x^{-2/3}. 3. Differentiate v(x) = x+1 v'(x) = 1. 4. Substitute into the Product Rule f'(x) = \left(\frac{1}{3}x^{-2/3}\right)(x+1) + (x^{1/3})(1).

Explanation

1. Apply the Product Rule<br /> The derivative of $f(x) = u(x)v(x)$ is $f'(x) = u'(x)v(x) + u(x)v'(x)$. Here, $u(x) = \sqrt[3]{x}$ and $v(x) = x+1$.<br />2. Differentiate $u(x) = \sqrt[3]{x}$<br /> $u(x) = x^{1/3}$, so $u'(x) = \frac{1}{3}x^{-2/3}$.<br />3. Differentiate $v(x) = x+1$<br /> $v'(x) = 1$.<br />4. Substitute into the Product Rule<br /> $f'(x) = \left(\frac{1}{3}x^{-2/3}\right)(x+1) + (x^{1/3})(1)$.
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