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Worked Examples · Example 3

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

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✓ Free question

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

Steps 2-3 — difference quotient, then rationalise:

x+h−xh=x+h−xh⋅x+h+xx+h+x=(x+h)−xh(x+h+x)=hh(x+h+x)=1x+h+x.\frac{\sqrt{x+h}-\sqrt{x}}{h}=\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}=\frac{(x+h)-x}{h\left(\sqrt{x+h}+\sqrt{x}\right)}=\frac{h}{h\left(\sqrt{x+h}+\sqrt{x}\right)}=\frac{1}{\sqrt{x+h}+\sqrt{x}}.

Step 4 — limit:

f′(x)=lim⁡h→01x+h+x=1x+x=12x.f'(x)=\lim_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{\sqrt{x}+\sqrt{x}}=\frac{1}{2\sqrt{x}}.

Verify (power rule): x=x1/2\sqrt{x}=x^{1/2}, so ddxx1/2=12x−1/2=12x\frac{d}{dx}x^{1/2}=\tfrac12 x^{-1/2}=\dfrac{1}{2\sqrt{x}}. Agrees.

✓Final answer

f′(x)=12xf'(x)=\dfrac{1}{2\sqrt{x}}.

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