#x*y = a(y+√(y²-x²))# then show that #x³*dy/dx=y²(y+√(y²-x²))# ?

Question

Aurora Alvarado · Accepted Answer

#dy/dx = (y^2)/(x^3)(y+sqrt(y^2-y^2))#

Making #f(x,y) = (xy)/(y+sqrt(y^2-x^2))-a=0#

we have

#df = f_x dx + f_y dy = 0#

with

#f_x = y^2/(-x^2 + y (y + sqrt[-x^2 + y^2]))# #f_y = (x (y - sqrt[-x^2 + y^2]))/(x^2 - y (y + sqrt[-x^2 + y^2]))#

so

#dy/dx = -(f_x)/(f_y) = (y^2)/(x^3)(y+sqrt(y^2-y^2))#

Isaiah Allen · Answer

undefined To show that $ x^3 \frac{dy}{dx} = y^2(y + \sqrt{y^2 - x^2}) $, differentiate both sides of the given equation $ xy = a(y + \sqrt{y^2 - x^2}) $ with respect to $ x $ implicitly, then manipulate the resulting expression to obtain the desired form.

First, differentiate both sides of the equation $ xy = a(y + \sqrt{y^2 - x^2}) $ with respect to $ x $ using the product rule and the chain rule:

$$ \frac{d}{dx}(xy) = \frac{d}{dx}(a(y + \sqrt{y^2 - x^2})) $$

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

$$ y + x\frac{dy}{dx} = a\frac{dy}{dx} + \frac{ax}{\sqrt{y^2 - x^2}} $$

Now, solve for $ \frac{dy}{dx} $:

$$ x\frac{dy}{dx} - a\frac{dy}{dx} = a\frac{ax}{\sqrt{y^2 - x^2}} - y $$

$$ (x - a)\frac{dy}{dx} = \frac{ax}{\sqrt{y^2 - x^2}} - y $$

$$ \frac{dy}{dx} = \frac{ax}{(x - a)\sqrt{y^2 - x^2}} - \frac{y}{x - a} $$

Now, multiply both sides by $ x^3 $:

$$ x^3 \frac{dy}{dx} = x^3\left(\frac{ax}{(x - a)\sqrt{y^2 - x^2}} - \frac{y}{x - a}\right) $$

$$ x^3 \frac{dy}{dx} = \frac{ax^4}{(x - a)\sqrt{y^2 - x^2}} - \frac{x^3y}{x - a} $$

$$ x^3 \frac{dy}{dx} = \frac{ax^4}{(x - a)\sqrt{y^2 - x^2}} - \frac{x^3y - ax^3}{x - a} $$

$$ x^3 \frac{dy}{dx} = \frac{ax^4}{(x - a)\sqrt{y^2 - x^2}} - \frac{ax^3}{x - a} $$

$$ x^3 \frac{dy}{dx} = \frac{ax^4 - ax^3\sqrt{y^2 - x^2}}{(x - a)\sqrt{y^2 - x^2}} $$

$$ x^3 \frac{dy}{dx} = \frac{ax^3(x - \sqrt{y^2 - x^2})}{(x - a)\sqrt{y^2 - x^2}} $$

$$ x^3 \frac{dy}{dx} = \frac{ax^3(y + \sqrt{y^2 - x^2})}{(x - a)\sqrt{y^2 - x^2}} $$

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

Thus, we have shown that $ x^3 \frac{dy}{dx} = y^2(y + \sqrt{y^2 - x^2}) $.