Math · Calculus I · Worked example

Use the power rule on roots and fractions

Differentiate f(x) = 4√x − 3/x² + 5, then find the slope of its graph at x = 4.

f (x) = 4 \sqrt{x} - \frac{3}{x^{2}} + 5

Rewrite each term as a power

√x is x^(1/2) and 3/x² is 3x^(−2). The constant 5 stays as it is. The function is defined for x > 0, because of the square root and the division by x².

f (x) = 4 x^{1 / 2} - 3 x^{- 2} + 5

Apply the power rule term by term

Bring down each exponent and lower it by one. For 4x^(1/2) that gives 4·(1/2)x^(−1/2) = 2x^(−1/2). For −3x^(−2) it gives −3·(−2)x^(−3) = 6x^(−3). The constant 5 contributes 0.

f^{'} (x) = 2 x^{- 1 / 2} + 6 x^{- 3}

Write the answer without negative exponents

x^(−1/2) is 1/√x and x^(−3) is 1/x³. Both forms are correct; this one is easier to evaluate by hand.

f^{'} (x) = \frac{2}{\sqrt{x}} + \frac{6}{x^{3}}

Evaluate the slope at x = 4

√4 = 2 and 4³ = 64, so f′(4) = 2/2 + 6/64 = 1 + 3/32 = 35/32 ≈ 1.094. The graph is rising at x = 4, a little more steeply than a 45° line.

f^{'} (4) = \frac{2}{2} + \frac{6}{64} = \frac{35}{32}

Result

f′(x) = 2/√x + 6/x³, and the slope at x = 4 is f′(4) = 35/32 ≈ 1.094.

f^{'} (x) = \frac{2}{\sqrt{x}} + \frac{6}{x^{3}}

Your turn

Differentiate f(x) = x³ − 6x² + 9x and find f′(2).

Show the answer and explanation

f′(x) = 3x² − 12x + 9, and f′(2) = −3.

Differentiate term by term with the power rule: 3x² − 12x + 9. Then f′(2) = 12 − 24 + 9 = −3, so the graph is falling at x = 2.

\begin{array}{l} f (x) = x^{3} - 6 x^{2} + 9 x \\ f^{'} (x) = 3 x^{2} - 12 x + 9 \\ f^{'} (2) = - 3 \end{array}

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Check the derivative in the Derivative & antiderivative checker, then change 4√x to 4∛x and predict the new derivative first.

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