Let f(x) = √[4]x = x^1/4. Find f'(16).

Apply the power rule to f(x) = x^1/4: bring down the exponent and subtract 1 from it, giving f'(x) = 1/4x^1/4 - 1 = 1/4x^-3/4. Now substitute x = 16: f'(16) = 1/4(16)^-3/4. Since 16^1/4 = 2, we have 16^3/4 = 2^3 = 8, so f'(16) = 1/4cdot1/8 = 1/32.

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Derivative of a fractional power function evaluated at a point

Let $f(x) = \sqrt[4]{x} = x^{1/4}$ . Find $f'(16)$ .

Hint

Use the power rule $\frac{d}{dx}[x^n] = n x^{n-1}$ with $n = 1/4$ , then substitute $x = 16$ into the resulting derivative formula.

Step-by-step solution

Apply the power rule to $f(x) = x^{1/4}$ : bring down the exponent and subtract 1 from it, giving $f'(x) = \frac{1}{4}x^{1/4 - 1} = \frac{1}{4}x^{-3/4}$ .

Now substitute $x = 16$ : $f'(16) = \frac{1}{4}(16)^{-3/4}$ . Since $16^{1/4} = 2$ , we have $16^{3/4} = 2^3 = 8$ , so $f'(16) = \frac{1}{4}\cdot\frac{1}{8} = \frac{1}{32}$ .

Derivative of a fractional power function evaluated at a point

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