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Given f(x) = (1)/(x), then f(x+h) = (1)/(x+h).

Mathematics

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Answer

f′(x)=−1x2f'(x) = -\frac{1}{x^2}

Graph of f(x) = 1/x

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Key points of f(x) = 1/x
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Step-by-step solution

5 steps

To find the derivative of y=1xy = \frac{1}{x} from first principles, we use the definition:

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

  1. Identify f(x)f(x) and f(x+h)f(x+h).

    Given f(x)=1xf(x) = \frac{1}{x}, then f(x+h)=1x+hf(x+h) = \frac{1}{x+h}.

  2. Substitute f(x)f(x) and f(x+h)f(x+h) into the first principles formula.

    f′(x)=lim⁡h→01x+h−1xhf'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}

  3. Simplify the numerator by finding a common denominator.

    f′(x)=lim⁡h→0x−(x+h)x(x+h)hf'(x) = \lim_{h \to 0} \frac{\frac{x - (x+h)}{x(x+h)}}{h} f′(x)=lim⁡h→0x−x−hx(x+h)hf'(x) = \lim_{h \to 0} \frac{\frac{x - x - h}{x(x+h)}}{h} f′(x)=lim⁡h→0−hx(x+h)hf'(x) = \lim_{h \to 0} \frac{\frac{-h}{x(x+h)}}{h}

  4. Simplify the complex fraction.

    f′(x)=lim⁡h→0−hx(x+h)⋅1hf'(x) = \lim_{h \to 0} \frac{-h}{x(x+h)} \cdot \frac{1}{h} f′(x)=lim⁡h→0−1x(x+h)f'(x) = \lim_{h \to 0} \frac{-1}{x(x+h)}

  5. Evaluate the limit as h→0h \to 0.

    Substitute h=0h=0 into the expression.

    f′(x)=−1x(x+0)f'(x) = \frac{-1}{x(x+0)} f′(x)=−1x(x)f'(x) = \frac{-1}{x(x)} f′(x)=−1x2f'(x) = \boxed{-\frac{1}{x^2}}
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