Power Rule and Basic Rules

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Key Ideas Easy Medium Hard

\(\dfrac{d}{dx} x^n = nx^{n-1}\)

What to Remember

Power rule

\(\dfrac{d}{dx} x^n = n x^{n - 1}\)

The exponent can be negative or a fraction. Rewrite a root or a denominator as a power first.

Sum, difference, and constant multiple

\(\dfrac{d}{dx}\bigl(cf(x)\bigr) = c\, f'(x)\)

\(\dfrac{d}{dx}\bigl(f(x) + g(x)\bigr) = f'(x) + g'(x)\)

The derivative of a constant is 0.

Expand a product before this rule

On this page, multiply a product out and write a quotient as a sum of powers. The product and quotient rules come next.

Then substitute

If the question asks for \(f'(a)\), find \(f'(x)\) first. Substitute a on the next line.

Level 1

Easy

Polynomials. One term at a time.

Example

Find \(f'(x)\) for \(f(x) = 5x^4 - 3x^2 + 7x - 2\).

\(\dfrac{d}{dx} x^n = n x^{n - 1}\)

\(\dfrac{d}{dx}(5x^4) = 20x^3\)

\(\dfrac{d}{dx}(-3x^2) = -6x\)

\(\dfrac{d}{dx}(7x) = 7\)

\(\dfrac{d}{dx}(-2) = 0\)

\(f'(x) = 20x^3 - 6x + 7\)

Answer: \(20x^3 - 6x + 7\)

Find \(f'(x)\) for \(f(x) = 6x^5 + 4x\).

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\(\dfrac{d}{dx}(6x^5) = 30x^4\)

\(\dfrac{d}{dx}(4x) = 4\)

\(30x^4 + 4\)

Answer: \(30x^4 + 4\)

Find \(f'(x)\) for \(f(x) = -2x^3 + 9\).

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\(\dfrac{d}{dx}(-2x^3) = -6x^2\)

\(\dfrac{d}{dx}(9) = 0\)

\(-6x^2\)

Answer: \(-6x^2\)

Find \(f'(x)\) for \(f(x) = x^7 - x\).

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\(\dfrac{d}{dx}(x^7) = 7x^6\)

\(\dfrac{d}{dx}(x) = 1\)

\(7x^6 - 1\)

Answer: \(7x^6 - 1\)

Find \(f'(x)\) for \(f(x) = 8\).

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The derivative of a constant is 0.

\(0\)

Answer: \(0\)

Find \(f'(x)\) for \(f(x) = \dfrac{1}{2}x^6 - 4x^3\).

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\(\dfrac{1}{2} \cdot 6x^5 = 3x^5\)

\(\dfrac{d}{dx}(-4x^3) = -12x^2\)

\(3x^5 - 12x^2\)

Answer: \(3x^5 - 12x^2\)

Level 2

Medium

Rewrite a fraction or a root as a power. Then use the power rule.

Example

Find \(f'(x)\) for \(f(x) = \dfrac{4}{x^2}\).

\(f(x) = 4x^{-2}\)

\(f'(x) = 4 \cdot (-2) x^{-3}\)

\(-8x^{-3}\)

\(-\dfrac{8}{x^3}\)

Answer: \(-\dfrac{8}{x^3}\)

Find \(f'(x)\) for \(f(x) = 3\sqrt{x}\).

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\(f(x) = 3x^{1/2}\)

\(f'(x) = 3 \cdot \dfrac{1}{2} x^{-1/2}\)

\(\dfrac{3}{2\sqrt{x}}\)

Answer: \(\dfrac{3}{2\sqrt{x}}\)

Find \(f'(x)\) for \(f(x) = x^2 + \dfrac{1}{x}\).

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\(f(x) = x^2 + x^{-1}\)

\(f'(x) = 2x + (-1)x^{-2}\)

\(2x - \dfrac{1}{x^2}\)

Answer: \(2x - \dfrac{1}{x^2}\)

Find \(f'(x)\) for \(f(x) = \dfrac{5}{x} - 2\sqrt{x}\).

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\(f(x) = 5x^{-1} - 2x^{1/2}\)

\(\dfrac{d}{dx}(5x^{-1}) = -5x^{-2}\)

\(\dfrac{d}{dx}(2x^{1/2}) = x^{-1/2}\)

\(-\dfrac{5}{x^2} - \dfrac{1}{\sqrt{x}}\)

Answer: \(-\dfrac{5}{x^2} - \dfrac{1}{\sqrt{x}}\)

Find \(f'(x)\) for \(f(x) = (x^2 + 3)(x - 4)\) by expanding first.

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\(f(x) = x^3 - 4x^2 + 3x - 12\)

\(f'(x) = 3x^2 - 8x + 3\)

Answer: \(3x^2 - 8x + 3\)

For \(f(x) = 3x^4 - 5x\), find \(f'(2)\).

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\(f'(x) = 12x^3 - 5\)

Substitute \(x = 2\).

\(f'(2) = 12 \cdot 8 - 5\)

\(96 - 5 = 91\)

Answer: \(91\)

Level 3

Hard

A rational exponent, or a quotient rewritten as a sum.

Example

Find \(f'(x)\) for \(f(x) = 7x^{4/3}\).

\(\dfrac{d}{dx} x^n = n x^{n - 1}\)

\(f'(x) = 7 \cdot \dfrac{4}{3} x^{1/3}\)

\(\dfrac{28}{3} x^{1/3}\)

Answer: \(\dfrac{28}{3} x^{1/3}\)

Find \(f'(x)\) for \(f(x) = \dfrac{2}{x^3} - \dfrac{6}{x}\).

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\(f(x) = 2x^{-3} - 6x^{-1}\)

\(\dfrac{d}{dx}(2x^{-3}) = -6x^{-4}\)

\(\dfrac{d}{dx}(-6x^{-1}) = 6x^{-2}\)

\(-\dfrac{6}{x^4} + \dfrac{6}{x^2}\)

Answer: \(-\dfrac{6}{x^4} + \dfrac{6}{x^2}\)

Find \(f'(x)\) for \(f(x) = \dfrac{x^2 - 1}{x}\), where \(x \neq 0\).

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\(f(x) = x - \dfrac{1}{x} = x - x^{-1}\)

\(f'(x) = 1 + x^{-2}\)

\(1 + \dfrac{1}{x^2}\)

Answer: \(1 + \dfrac{1}{x^2}\)

Find \(f'(x)\) for \(f(x) = \dfrac{x^2 + 4x}{x}\), where \(x \neq 0\).

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\(f(x) = x + 4\)

\(f'(x) = 1\)

Answer: \(1\)

Which expression is the derivative of \(x^{-3} + 4x\)?

A. \(-3x^{-4} + 4\)

B. \(-3x^{-2} + 4\)

C. \(x^{-4} + 4\)

D. \(-3x^{-4} + 4x\)

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\(\dfrac{d}{dx} x^{-3} = -3x^{-4}\)

\(\dfrac{d}{dx}(4x) = 4\)

\(-3x^{-4} + 4\)

Answer: A

For \(f(x) = 4\sqrt{x} + \dfrac{1}{x}\), find \(f'(4)\).

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

\(f'(x) = 2x^{-1/2} - x^{-2}\)

\(f'(x) = \dfrac{2}{\sqrt{x}} - \dfrac{1}{x^2}\)

Substitute \(x = 4\).

\(\dfrac{2}{2} - \dfrac{1}{16} = 1 - \dfrac{1}{16}\)

\(\dfrac{15}{16}\)

Answer: \(\dfrac{15}{16}\)

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