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Question Number 204227 by abdomath last updated on 09/Feb/24

Answered by deleteduser1 last updated on 09/Feb/24

$$Lety=f\left(x\right)\Rightarrow x^2{y}^3={(x+y)}^5\Rightarrow x^2{\left[f\left(x\right)\right]}^3={(x+f\left(x\right))}^5$$$$\Rightarrow \frac{d}{dx}\left(x^2{\left[f\left(x\right)\right]}^3\right)=\frac{d}{dx}{(x+f\left(x\right))}^5$$$$\Rightarrow 3x^2{\left[f\left(x\right)\right]}^2{f}^{\prime }\left(x\right)+2x{\left[f\left(x\right)\right]}^3=5{(x+f\left(x\right))}^4(1+f^{\prime }\left(x\right))$$$$\Rightarrow f^{\prime }\left(x\right)[3x^2{y}^2-5{(x+y)}^4]=5{(x+y)}^4-2{xy}^3$$$$\Rightarrow \frac{dy}{dx}=\frac{5{(x+y)}^4-2{xy}^3}{3x^2{y}^2-5{(x+y)}^4}$$

Answered by cortano12 last updated on 09/Feb/24

$$\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^2{\mathrm{y}}^3\right)=\frac{\mathrm{d}}{\mathrm{dx}}{(\mathrm{x}+\mathrm{y})}^5$$$${2\mathrm{xy}}^3+{3\mathrm{x}}^2{\mathrm{y}}^2{\mathrm{y}}^{\prime }=5{(\mathrm{x}+\mathrm{y})}^4(1+{\mathrm{y}}^{\prime })$$$${3\mathrm{x}}^2{\mathrm{y}}^2{\mathrm{y}}^{\prime }-5{(\mathrm{x}+\mathrm{y})}^4{\mathrm{y}}^{\prime }=5{(\mathrm{x}+\mathrm{y})}^4-{2\mathrm{xy}}^3$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{5{(\mathrm{x}+\mathrm{y})}^4-{2\mathrm{xy}}^3}{{3\mathrm{x}}^2{\mathrm{y}}^2-5{(\mathrm{x}+\mathrm{y})}^4}$$

Answered by Frix last updated on 09/Feb/24

$$2{xy}^{3}dx+3x^2{y}^{2}dy=5{(x+y)}^{4}dx+5{(x+y)}^{4}dy$$$$(3x^2{y}^2-5{(x+y)}^4)dy=(5{(x+y)}^4-2{xy}^3)dx$$$$\frac{dy}{dx}=\frac{5{(x+y)}^4-2{xy}^3}{3x^2{y}^2-5{(x+y)}^4}$$

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