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Worked Examples · Example 2

Q.Differentiate f(x)=1xf(x) = \dfrac{1}{x} from first principles.

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✓ Free question

Step 1: f(x+h)=1x+hf(x+h)=\dfrac{1}{x+h}.

Step 2 — subtract over a common denominator:

f(x+h)−f(x)=1x+h−1x=x−(x+h)x(x+h)=−hx(x+h).f(x+h)-f(x)=\frac{1}{x+h}-\frac{1}{x}=\frac{x-(x+h)}{x(x+h)}=\frac{-h}{x(x+h)}.

Step 3 — divide by hh:

f(x+h)−f(x)h=−hx(x+h)⋅1h=−1x(x+h).\frac{f(x+h)-f(x)}{h}=\frac{-h}{x(x+h)}\cdot\frac{1}{h}=\frac{-1}{x(x+h)}.

Step 4 — limit:

f′(x)=lim⁡h→0−1x(x+h)=−1x⋅x=−1x2.f'(x)=\lim_{h\to0}\frac{-1}{x(x+h)}=\frac{-1}{x\cdot x}=-\frac{1}{x^2}.

Verify (power rule): 1x=x−1\frac1x=x^{-1}, so ddxx−1=−1⋅x−2=−1x2\frac{d}{dx}x^{-1}=-1\cdot x^{-2}=-\frac{1}{x^2}. Agrees.

✓Final answer

f′(x)=−1x2f'(x)=-\dfrac{1}{x^{2}}.

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