How do you implicitly differentiate $y$ from $y\sqrt{x^2+y^2} = 15$?

I've been working on this problem for the last 45 minutes, and I keep getting the wrong answer, no matter what I do. I tried squaring the whole equation, so that there was no radical to deal with - didn't work. I tried the product rule + chain rule - didn't work. Is anyone able to explain how to do it? I'm really desperate.

Thank you so much!

• Hint: You may begin by noting that $$\sqrt{x^2+y^2}=15\qquad\iff\qquad x^2+y^2=225$$ – Ángel Mario Gallegos Nov 3 '15 at 3:14

$$(x^2+y^2)^\frac 12 = 15$$ differentiating ... $$(x^2+y^2)^{-\frac 12}(2x+2yy') = 0$$ clearly only the second factor can be zero so

$$y'=-\frac xy$$

as required for a circle (slope of tangent is the negative reciprocal of the slope of the radius )

• I thought the expression is $y (x^2+y^2)^\frac 12 = 15$ – ja72 Nov 3 '15 at 6:39
• @ja72 The question was changed after I posted my answer – WW1 Nov 3 '15 at 6:44

Assume $y(x)$ is a function of $x$ on an open domain around $(x_0,y_0)$, then: $$0=\left(y(x)\sqrt{y(x)^2+x^2}-15\right)'=y'(x)\sqrt{y(x)^2+x^2}+y(x)\frac{y(x)y'(x)+x}{\sqrt{y(x)^2+x^2}}\Leftrightarrow$$ $$0=y'(x)\left(2y(x)^2+x^2\right)+xy(x)\Rightarrow y'(x)=-\frac{xy(x)}{2y(x)^2+x^2}$$ Meaning: $$y'=-\frac{xy(x)}{2y^2+x^2}$$

You have $f(x,y)=C$ and apply the chain rule

$$\frac{\partial f(x,y)}{\partial x} {\rm d}x + \frac{\partial f(x,y)}{\partial y} {\rm d} y = 0$$

$$\frac{x y}{\sqrt{x^2+y^2}} {\rm d}x + \frac{x^2+2 y^2}{\sqrt{x^2+y^2}} {\rm d} y = 0$$
$$\frac{{\rm d}y}{{\rm d}x} = - \frac{x y}{x^2+2 y^2}$$