Question

If $y=\frac{x}{x+2}$ then $x\frac{dy}{dx}=$?

• A. $1$
• B. $(1-y)$
• C. y
• D. $(1-y)y$

Answer 1

Answer: (D)

$(1-y)y$

Q. If $y=\frac{x}{x+2}$ then $x\frac{dy}{dx}=?$

Sol${}^n\Rightarrow$ We have $y=\frac{x}{x+2}$

So, $y(x+2)-x=0$

$(y-1)x=-2y$

$x=\frac{-2y}{y-1}=\frac{2y}{1-y}$

now, differentiate y wrt x we get

$\frac{dy}{dx}=\frac{(x+2)\left(1\right)-\left(x\right)(1+0)}{(x+2{)}^2}$

$\frac{dy}{dx}=\frac{2}{(x+2{)}^2}$

multiply $x^2$ on both side we gets

$x^2\frac{dy}{dx}=\frac{2x^2}{(x+2{)}^2}$

$x\left(\frac{2y}{1-y}\right)\frac{dy}{dx}=2y^2$

$x\frac{dy}{dx}=\frac{2y^2}{\left(\frac{2y}{1-y}\right)}$

$x\frac{dy}{dx}=y(1-y)$

So, $x\frac{dy}{dx}=(1-y)y$