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Question Number 13068 by tawa tawa last updated on 13/May/17

$$\mathrm{If}\mathrm{y}=\sqrt[3]{\mathrm{x}}\mathrm{Find}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{from}\mathrm{the}\mathrm{first}\mathrm{principle}$$

Answered by ajfour last updated on 13/May/17

$$\frac{dy}{dx}=\underset{h\to 0}{\lim }\frac{{(x+h)}^{1/3}-x^{1/3}}{h}$$$$=\underset{h\to 0}{\lim }\frac{x^{1/3}\{{(1+\frac{h}{x})}^{1/3}-1\}}{h}$$$$ash\ll xwehave{(1+\frac{h}{x})}^n=1+\frac{nh}{x}$$$$so\frac{dy}{dx}=\underset{h\to 0}{\lim }\frac{x^{1/3}\{1+\frac{h}{3x}-1\}}{h}$$$$\Rightarrow \frac{dy}{dx}=\left(x^{1/3}\right)\underset{h\to 0}{\lim }\frac{\left(h/3x\right)}{h}$$$$=x^{1/3}\left(\frac{1}{3x}\right)$$$$=\frac{1}{3x^{2/3}}.$$

Commented by tawa tawa last updated on 13/May/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Nayon last updated on 23/May/17

$$why{(1+\frac{h}{x})}^n=1+\frac{nh}{x}?$$

Answered by mrW1 last updated on 13/May/17

$$Generallyfory=f\left(x\right)=x^a(a\ne 0)wehave$$$$f(x+\mathrm{\Delta }x)=f(x+h)={(x+h)}^a$$$$\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{f(x+\mathrm{\Delta }x)-f\left(x\right)}{\mathrm{\Delta }x}=\frac{{(x+h)}^a-x^a}{h}=x^a\frac{{(1+\frac{h}{x})}^a-1}{h}$$$$\frac{dy}{dx}=\underset{h\to 0}{\lim }\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=x^a\times \underset{h\to 0}{\lim }\frac{{(1+\frac{h}{x})}^a-1}{h}$$$${(1+\frac{h}{x})}^a=1+a\left(\frac{h}{x}\right)+\frac{a(a-1)}{2!}{\left(\frac{h}{x}\right)}^2+\frac{a(a-1)(a-2)}{3!}{\left(\frac{h}{x}\right)}^3+\cdot \cdot \cdot$$$$\frac{{(1+\frac{h}{x})}^a-1}{h}=\frac{a}{x}+\frac{a(a-1)h}{2!x^2}+\frac{a(a-1)(a-2)h^2}{3!x^3}+\cdot \cdot \cdot$$$$\underset{h\to 0}{\lim }\frac{{(1+\frac{h}{x})}^a-1}{h}=\frac{a}{x}+0+0+0+\cdot \cdot \cdot =\frac{a}{x}$$$$\Rightarrow \frac{dy}{dx}=x^a\times \frac{a}{x}={ax}^{a-1}$$$$fora=\frac{1}{3}weget$$$$\frac{dy}{dx}=\frac{1}{3}{x}^{-\frac{2}{3}}$$

Commented by ajfour last updated on 13/May/17

$$toogood!$$

Commented by tawa tawa last updated on 13/May/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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