Question

Find the derivative of the following function by the first principle: ${\displaystyle x\sqrt{x}}$

Answer 1

Let f(x) = ${\displaystyle x\sqrt{x}=x^{\frac{3}{2}}}$

∴ f(x + h) = ${\displaystyle {(x+\text{h})}^{\frac{3}{2}}}$

By first principle, we get

f ′(x) =${\displaystyle \underset{\text{h}\to 0}{lim}\frac{\text{f}(x+\text{h})-\text{f}\left(x\right)}{\text{h}}}$

=${\displaystyle \underset{\text{h}\to 0}{lim}\frac{{(x+\text{h})}^{\frac{3}{2}}-x^{\frac{3}{2}}}{\text{h}}}$

=${\displaystyle \underset{\text{h}\to 0}{lim}\frac{[{(x+\text{h})}^{\frac{3}{2}}-x^{\frac{3}{2}}][{(x+h)}^{\frac{3}{2}}+x^{\frac{3}{2}}]}{h[{(x+h)}^{\frac{3}{2}}+x^{\frac{3}{2}}]}}$

=${\displaystyle \underset{\text{h}\to 0}{lim}\frac{{(x+\text{h})}^3-x^3}{\text{h}[{(x+\text{h})}^{\frac{3}{2}}+x^{\frac{3}{2}}]}}$

=${\displaystyle \underset{\text{h}\to 0}{lim}\frac{x^3+3x^2\text{h}+3x{\text{h}}^2+{\text{h}}^3-x^3}{\text{h}[{(x+\text{h})}^{\frac{3}{2}}+x^{\frac{3}{2}})}}$

= ${\displaystyle \underset{\text{h}\to 0}{lim}\frac{\text{h}(3x^2+3x\text{h}+{\text{h}}^2)}{\text{h}[{(x+\text{h})}^{\frac{3}{2}}+x^{\frac{3}{2}}]}}$

= ${\displaystyle \underset{\text{h}\to 0}{lim}\frac{\text{h}(3x^2+3x\text{h}+{\text{h}}^2)}{\text{h}[{(x+\text{h})}^{\frac{3}{2}}+x^{\frac{3}{2}}]}}$

= ${\displaystyle \underset{\text{h}\to 0}{lim}\frac{3x^2+3x\text{h}+{\text{h}}^2}{{(x+\text{h})}^{\frac{3}{2}}+ x^{\frac{3}{2}}}}$  ...[∵ h → 0, ∴ h ≠ 0]

= ${\displaystyle \frac{3x^2+3x\times 0+0^2}{{(x+0)}^{\frac{3}{2}}+x^{\frac{3}{2}}}}$

= ${\displaystyle \frac{3x^2}{2x^{\frac{3}{2}}}}$

=${\displaystyle \frac{3}{2}{x}^{\frac{1}{2}}}$

= ${\displaystyle \frac{3}{2}\sqrt{x}}$