Find-dy-dx-x-2-3-6-x-1-3-to-it-simplest-form-

Question Number 28280 by Cheyboy last updated on 23/Jan/18

$${Find}\:{dy}/{dx} \\ $$$${x}^{\frac{\mathrm{2}}{\mathrm{3}}} \left(\mathrm{6}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}\:}} \:{to}\:{it}\:{simplest}\:{form} \\ $$

Commented by abdo imad last updated on 23/Jan/18

we have y(x)=(x^2 (6−x))^(1/3) =(−x^3 +6x^2 )^(1/3) so (dy/dx)(x)= (1/3)( −3x^2 +12x)^(−(2/3)) .

$${we}\:{have}\:{y}\left({x}\right)=\left({x}^{\mathrm{2}} \left(\mathrm{6}−{x}\right)\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\left(−{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:{so} \\ $$$$\frac{{dy}}{{dx}}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\:−\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{12}{x}\right)^{−\frac{\mathrm{2}}{\mathrm{3}}} \:\:. \\ $$

Commented by Cheyboy last updated on 23/Jan/18

$${sir}\:{d}\:{book}\:{is}\:{having}\:{different}\:{thing} \\ $$$$ \\ $$

Commented by abdo imad last updated on 23/Jan/18

$${trust}\:{your}\:{self}\:{sir}\:{because}\:{a}\:{lots}\:{of}\:{books}\:{are}\:{full}\:{with}\:{error} \\ $$$${and}\:{perhaps}\:{the}\:{book}\:{have}\:{given}\:{another}\:{form}\:{of}\:{derivative} \\ $$$${equal}\:{to}\:{this}\:\:{my}\:{answer}\:{is}\:{correct}… \\ $$

Commented by Cheyboy last updated on 24/Jan/18

$${ok}\:{thankz}\:{sir} \\ $$$$ \\ $$

Answered by ajfour last updated on 24/Jan/18

y=x^(2/3) (6−x)^(1/3) ln y=(2/3)ln x+(1/3)ln (6−x) (1/y)(dy/dx)=(2/(3x))−(1/(3(6−x))) (dy/dx)=y[((12−2x−x)/(3x(6−x)))] (dy/dx)=((x^(2/3) (6−x)^(1/3) (4−x))/(x(6−x))) =((4−x)/(x^(1/3) (6−x)^(2/3) )) .

$${y}={x}^{\mathrm{2}/\mathrm{3}} \left(\mathrm{6}−{x}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\mathrm{ln}\:{y}=\frac{\mathrm{2}}{\mathrm{3}}\mathrm{ln}\:{x}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left(\mathrm{6}−{x}\right) \\ $$$$\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dx}}=\frac{\mathrm{2}}{\mathrm{3}{x}}−\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{6}−{x}\right)} \\ $$$$\:\:\:\frac{{dy}}{{dx}}={y}\left[\frac{\mathrm{12}−\mathrm{2}{x}−{x}}{\mathrm{3}{x}\left(\mathrm{6}−{x}\right)}\right] \\ $$$$\frac{{dy}}{{dx}}=\frac{{x}^{\mathrm{2}/\mathrm{3}} \left(\mathrm{6}−{x}\right)^{\mathrm{1}/\mathrm{3}} \left(\mathrm{4}−{x}\right)}{{x}\left(\mathrm{6}−{x}\right)} \\ $$$$\:\:\:\:\:=\frac{\mathrm{4}−{x}}{{x}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{6}−{x}\right)^{\mathrm{2}/\mathrm{3}} }\:. \\ $$

Commented by Cheyboy last updated on 24/Jan/18