Question Number 204227 by abdomath last updated on 09/Feb/24
Answered by AST last updated on 09/Feb/24
![Let y=f(x)⇒x^2 y^3 =(x+y)^5 ⇒x^2 [f(x)]^3 =(x+f(x))^5 ⇒(d/dx)(x^2 [f(x)]^3 )=(d/dx)(x+f(x))^5 ⇒3x^2 [f(x)]^2 f′(x)+2x[f(x)]^3 =5(x+f(x))^4 (1+f′(x)) ⇒f′(x)[3x^2 y^2 −5(x+y)^4 ]=5(x+y)^4 −2xy^3 ⇒(dy/dx)=((5(x+y)^4 −2xy^3 )/(3x^2 y^2 −5(x+y)^4 ))](https://www.tinkutara.com/question/Q204228.png)
$${Let}\:{y}={f}\left({x}\right)\Rightarrow{x}^{\mathrm{2}} {y}^{\mathrm{3}} =\left({x}+{y}\right)^{\mathrm{5}} \Rightarrow{x}^{\mathrm{2}} \left[{f}\left({x}\right)\right]^{\mathrm{3}} =\left({x}+{f}\left({x}\right)\right)^{\mathrm{5}} \\ $$$$\Rightarrow\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} \left[{f}\left({x}\right)\right]^{\mathrm{3}} \right)=\frac{{d}}{{dx}}\left({x}+{f}\left({x}\right)\right)^{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{2}} \left[{f}\left({x}\right)\right]^{\mathrm{2}} {f}'\left({x}\right)+\mathrm{2}{x}\left[{f}\left({x}\right)\right]^{\mathrm{3}} =\mathrm{5}\left({x}+{f}\left({x}\right)\right)^{\mathrm{4}} \left(\mathrm{1}+{f}'\left({x}\right)\right) \\ $$$$\Rightarrow{f}'\left({x}\right)\left[\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{5}\left({x}+{y}\right)^{\mathrm{4}} \right]=\mathrm{5}\left({x}+{y}\right)^{\mathrm{4}} −\mathrm{2}{xy}^{\mathrm{3}} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{5}\left({x}+{y}\right)^{\mathrm{4}} −\mathrm{2}{xy}^{\mathrm{3}} }{\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{5}\left({x}+{y}\right)^{\mathrm{4}} } \\ $$
Answered by cortano12 last updated on 09/Feb/24
$$\:\frac{\mathrm{d}}{\mathrm{dx}}\:\left(\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{3}} \right)\:=\:\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{5}} \\ $$$$\:\mathrm{2xy}^{\mathrm{3}} +\:\mathrm{3x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:\mathrm{y}'\:=\:\mathrm{5}\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{4}} \left(\mathrm{1}+\mathrm{y}'\right) \\ $$$$\:\:\mathrm{3x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \mathrm{y}'−\mathrm{5}\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{4}} \mathrm{y}'=\:\mathrm{5}\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{4}} −\mathrm{2xy}^{\mathrm{3}} \\ $$$$\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{5}\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{4}} −\mathrm{2xy}^{\mathrm{3}} }{\mathrm{3x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} −\mathrm{5}\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{4}} } \\ $$
Answered by Frix last updated on 09/Feb/24
$$\mathrm{2}{xy}^{\mathrm{3}} {dx}+\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} {dy}=\mathrm{5}\left({x}+{y}\right)^{\mathrm{4}} {dx}+\mathrm{5}\left({x}+{y}\right)^{\mathrm{4}} {dy} \\ $$$$\left(\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{5}\left({x}+{y}\right)^{\mathrm{4}} \right){dy}=\left(\mathrm{5}\left({x}+{y}\right)^{\mathrm{4}} −\mathrm{2}{xy}^{\mathrm{3}} \right){dx} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{5}\left({x}+{y}\right)^{\mathrm{4}} −\mathrm{2}{xy}^{\mathrm{3}} }{\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{5}\left({x}+{y}\right)^{\mathrm{4}} } \\ $$