# Thread: Need help, quickly. Implicit differtiation

1. ## Need help, quickly. Implicit differtiation

I have been having trouble with this question. A few people have helped, but I've gotten no where from that. please help!

Question: Show that for the relation ((x/y)^.5) + ((y/x)^.5) = 10, x does not equal y which does not equal 0, dy/dx = y/x

Thanks

2. Hello, elle1019!

This is a messy one . . .

Show that the relation: .$\displaystyle \left(\frac{x}{y}\right)^{\frac{1}{2}} + \left(\frac{y}{x}\right)^{\frac{1}{2}} \:= \;10$

$\displaystyle \text{ where }x \neq y\,\text{ and }\,x,y \neq 0,\;\text{ produces: }\;\frac{dy}{dx} \:=\: \frac{y}{x}$
Differentiate:

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

. . . . . . . . $\displaystyle \frac{1}{2}\cdot\frac{y^{\frac{1}{2}}}{x^{\frac{1} {2}}}\cdot\frac{y-x\frac{dy}{dx}}{y^2} + \frac{1}{2}\cdot\frac{x^{\frac{1}{2}}}{y^{\frac{1} {2}}}\cdot\frac{x\frac{dy}{dx}-y}{x^2} \;=\;0$

. . . . . . . . . . . . . . . . . $\displaystyle \frac{y-x\frac{dy}{dx}}{2x^{\frac{1}{2}}y^{\frac{3}{2}}} + \frac{x\frac{dy}{dx}-y}{2x^{\frac{3}{2}}y^{\frac{1}{2}}} \;=\;0$

Multiply by $\displaystyle 2x^{\frac{3}{2}}y^{\frac{3}{2}}\!:\;\;x\left(y-x\frac{dy}{dx}\right) + y\left(x\frac{dy}{dx} - y\right) \;=\;0$

. . . . . . . . . . . . . . . . .$\displaystyle xy - x^2\frac{dy}{dx} + xy\frac{dy}{dx} - y^2 \;=\;0$

. . . . . . . . . . . . . . . . . . . . . . .$\displaystyle xy\frac{dy}{dx} - x^2\frac{dy}{dx} \;=\;y^2 - xy$

. . . . . . . . . . . . . . . . . . . . . . . .$\displaystyle x(y-x)\frac{dy}{dx} \;=\;y(y-x)$

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .$\displaystyle \frac{dy}{dx} \;=\;\frac{y(y-x)}{x(y-x)}$
. . Therefore: . $\displaystyle \frac{dy}{dx} \:=\:\frac{y}{x}$