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

If y = (x)^(1/3) Find (dy/dx) from the first principle

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

Answered by ajfour last updated on 13/May/17

(dy/dx)=lim_(h→0) (((x+h)^(1/3) −x^(1/3) )/h) =lim_(h→0) ((x^(1/3) {(1+(h/x))^(1/3) −1})/h) as h≪x we have (1+(h/x))^n =1+((nh)/x) so (dy/dx)=lim_(h→0) ((x^(1/3) {1+(h/(3x))−1})/h) ⇒(dy/dx)=(x^(1/3) )lim_(h→0) (((h/3x))/h) = x^(1/3) ((1/(3x))) = (1/(3x^(2/3) )) .

$$\frac{{dy}}{{dx}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+{h}\right)^{\mathrm{1}/\mathrm{3}} −{x}^{\mathrm{1}/\mathrm{3}} }{{h}} \\ $$$$\:\:\:\:=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{1}/\mathrm{3}} \left\{\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{\mathrm{1}/\mathrm{3}} −\mathrm{1}\right\}}{{h}} \\ $$$${as}\:{h}\ll{x}\:\:{we}\:{have}\:\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{n}} =\mathrm{1}+\frac{{nh}}{{x}} \\ $$$${so}\:\:\frac{{dy}}{{dx}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{1}/\mathrm{3}} \left\{\mathrm{1}+\frac{{h}}{\mathrm{3}{x}}−\mathrm{1}\right\}}{{h}} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\left({x}^{\mathrm{1}/\mathrm{3}} \right)\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({h}/\mathrm{3}{x}\right)}{{h}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:{x}^{\mathrm{1}/\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{3}{x}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{2}/\mathrm{3}} }\:. \\ $$

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

God bless you sir

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

Commented by Nayon last updated on 23/May/17

why(1+(h/x))^n =1+((nh)/x) ?

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

Answered by mrW1 last updated on 13/May/17

Generally for y=f(x)=x^a (a≠0) we have f(x+Δx)=f(x+h)=(x+h)^a ((Δy)/(Δx))=((f(x+Δx)−f(x))/(Δx))=(((x+h)^a −x^a )/h)=x^a (((1+(h/x))^a −1)/h) (dy/dx)=lim_(h→0) ((Δy)/(Δx))=x^a ×lim_(h→0) (((1+(h/x))^a −1)/h) (1+(h/x))^a =1+a((h/x))+((a(a−1))/(2!))((h/x))^2 +((a(a−1)(a−2))/(3!))((h/x))^3 +∙∙∙ (((1+(h/x))^a −1)/h)=(a/x)+((a(a−1)h)/(2!x^2 ))+((a(a−1)(a−2)h^2 )/(3!x^3 ))+∙∙∙ lim_(h→0) (((1+(h/x))^a −1)/h)=(a/x)+0+0+0+∙∙∙=(a/x) ⇒(dy/dx)=x^a ×(a/x)=ax^(a−1) for a=(1/3) we get (dy/dx)=(1/3)x^(−(2/3))

$${Generally}\:{for}\:{y}={f}\left({x}\right)={x}^{{a}} \:\left({a}\neq\mathrm{0}\right)\:{we}\:{have} \\ $$$${f}\left({x}+\Delta{x}\right)={f}\left({x}+{h}\right)=\left({x}+{h}\right)^{{a}} \\ $$$$\frac{\Delta{y}}{\Delta{x}}=\frac{{f}\left({x}+\Delta{x}\right)−{f}\left({x}\right)}{\Delta{x}}=\frac{\left({x}+{h}\right)^{{a}} −{x}^{{a}} }{{h}}={x}^{{a}} \frac{\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{a}} −\mathrm{1}}{{h}} \\ $$$$\frac{{dy}}{{dx}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\Delta{y}}{\Delta{x}}={x}^{{a}} ×\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{a}} −\mathrm{1}}{{h}} \\ $$$$\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{a}} =\mathrm{1}+{a}\left(\frac{{h}}{{x}}\right)+\frac{{a}\left({a}−\mathrm{1}\right)}{\mathrm{2}!}\left(\frac{{h}}{{x}}\right)^{\mathrm{2}} +\frac{{a}\left({a}−\mathrm{1}\right)\left({a}−\mathrm{2}\right)}{\mathrm{3}!}\left(\frac{{h}}{{x}}\right)^{\mathrm{3}} +\centerdot\centerdot\centerdot \\ $$$$\frac{\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{a}} −\mathrm{1}}{{h}}=\frac{{a}}{{x}}+\frac{{a}\left({a}−\mathrm{1}\right){h}}{\mathrm{2}!{x}^{\mathrm{2}} }+\frac{{a}\left({a}−\mathrm{1}\right)\left({a}−\mathrm{2}\right){h}^{\mathrm{2}} }{\mathrm{3}!{x}^{\mathrm{3}} }+\centerdot\centerdot\centerdot \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{a}} −\mathrm{1}}{{h}}=\frac{{a}}{{x}}+\mathrm{0}+\mathrm{0}+\mathrm{0}+\centerdot\centerdot\centerdot=\frac{{a}}{{x}} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}={x}^{{a}} ×\frac{{a}}{{x}}={ax}^{{a}−\mathrm{1}} \\ $$$${for}\:{a}=\frac{\mathrm{1}}{\mathrm{3}}\:{we}\:{get} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{3}}{x}^{−\frac{\mathrm{2}}{\mathrm{3}}} \\ $$

Commented by ajfour last updated on 13/May/17

too good !

$${too}\:{good}\:!\: \\ $$

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

God bless you sir.

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

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