Discussion Forum

If   $\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4$, then  $\frac{dy}{dx}=$

Answer:

Explanation:

We have

$\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4$

 squaring both sides , we get

$\frac{x}{y}+\frac{y}{x}+2=16$

           $x^{2}+y^{2}=14xy$

 On differentiating w.r.t 'x' , we get

$2x+2y\frac{dy}{dx}=14\left(x\frac{dy}{dx}+y\right)$

 $x+y\frac{dy}{dx}=7x\frac{dy}{dx}+7y$

$\frac{dy}{dx}(y-7x)=7y-x$

$\frac{dy}{dx}=\frac{7y-x}{y-7x}$