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Derivative of a fractional power function evaluated at a point

Let f(x)=x1/5f(x) = x^{1/5}. Use the power rule to find f′(x)f'(x), and then evaluate f′(32)f'(32).

Hint

Apply the power rule ddx[xn]=nxn−1\frac{d}{dx}[x^n] = n x^{n-1} with n=15n = \frac{1}{5}. Then substitute x=32x = 32, remembering that 32=2532 = 2^5.

Step-by-step solution

By the power rule, f′(x)=15x1/5−1=15x−4/5f'(x) = \frac{1}{5} x^{1/5 - 1} = \frac{1}{5} x^{-4/5}. This can also be written as f′(x)=15x4/5f'(x) = \dfrac{1}{5x^{4/5}}.

Now substitute x=32x = 32. Since 32=2532 = 2^5, we have 324/5=(25)4/5=24=1632^{4/5} = (2^5)^{4/5} = 2^4 = 16. So f′(32)=15⋅16=180f'(32) = \dfrac{1}{5 \cdot 16} = \dfrac{1}{80}.

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