Question

If $x^3{y}^5=(x+y{)}^8$, then show that $\frac{dy}{dx}=\frac{y}{x}$

Answer 1

$x^3{y}^5={(x+y)}^8$

$\Rightarrow x^3\left(5y^4\frac{dy}{dx}\right)+y^5\left(3x^2\right)=8{(x+y)}^7(1+\frac{dy}{dx})$

$\Rightarrow 5x^3{y}^4\frac{dy}{dx}+3x^2{y}^5=8{(x+y)}^7+8{(x+y)}^7\frac{dy}{dx}$

$\Rightarrow (5x^3{y}^4-8{(x+y)}^7)\frac{dy}{dx}=8{(x+y)}^7-3x^2{y}^5$

We have $x^3{y}^5={(x+y)}^8\Rightarrow x^3{y}^4=\frac{{(x+y)}^8}{y}$

and $x^2{y}^5={(x+y)}^8\Rightarrow x^2{y}^5=\frac{{(x+y)}^8}{x}$

$\Rightarrow (\frac{{5(x+y)}^8}{y}-8{(x+y)}^7)\frac{dy}{dx}=8{(x+y)}^7-\frac{{3(x+y)}^8}{x}$

$\Rightarrow {(x+y)}^7(\frac{5x+5y}{y}-8)\frac{dy}{dx}={(x+y)}^7(8-\frac{3x+3y}{x})$

$\Rightarrow \frac{5x+5y-8y}{y}\frac{dy}{dx}=\frac{8x-3x-3y}{x}$

$\Rightarrow \frac{5x-3y}{y}\frac{dy}{dx}=\frac{5x-3y}{x}$

$\therefore \frac{dy}{dx}=\frac{y}{x}$