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Question-21964




Question Number 21964 by chernoaguero@gmail.com last updated on 07/Oct/17
Answered by ibraheem160 last updated on 07/Oct/17
y=x^2 (2x−5)^4   (du/dx)=2x,  (dv/dx)=8(2x−5)^3      (dy/dx)=u(dv/dx)+v(du/dx)  ⇒x^2 .8(2x−5)^3 +2x(2x−5)^4   ⇒8x^2 (2x−5)^3 +2x(2x−5)^4   =2x(2x−5)^3 (6x−5)
$$\mathrm{y}=\mathrm{x}^{\mathrm{2}} \left(\mathrm{2x}−\mathrm{5}\right)^{\mathrm{4}} \\ $$$$\frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{2x},\:\:\frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{8}\left(\mathrm{2x}−\mathrm{5}\right)^{\mathrm{3}} \\ $$$$\: \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{u}\frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v}\frac{\mathrm{du}}{\mathrm{dx}} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} .\mathrm{8}\left(\mathrm{2x}−\mathrm{5}\right)^{\mathrm{3}} +\mathrm{2x}\left(\mathrm{2x}−\mathrm{5}\right)^{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{8x}^{\mathrm{2}} \left(\mathrm{2x}−\mathrm{5}\right)^{\mathrm{3}} +\mathrm{2x}\left(\mathrm{2x}−\mathrm{5}\right)^{\mathrm{4}} \\ $$$$=\mathrm{2x}\left(\mathrm{2x}−\mathrm{5}\right)^{\mathrm{3}} \left(\mathrm{6x}−\mathrm{5}\right) \\ $$
Commented by chernoaguero@gmail.com last updated on 08/Oct/17
Thank you sir
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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