1. ## Implicit differentiation.

Can you help with one more?
X^2+2xy+y^2=15
2x+2x(dy/dx)+2y+2y(dy/dx)=0

2. Alright... I'm not convinced I've got this right as I'm quite new to implicit differentiation myself... but here's what I got step by step (do not take this as correct though).

$\displaystyle \frac{d}{dx}(x^2) + \frac{d}{dx}(2xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(15)$

$\displaystyle 2x + 2y + 2x + \frac{dy}{dx}\frac{d}{dy}(y^2) = 0$

$\displaystyle 2x + 2y + 2x + \frac{dy}{dx}2y = 0$

$\displaystyle \frac{dy}{dx} = \frac{4x + 2y}{2y} = \frac{2x + y}{y}$

like I said, check it first, I'm havent convinced myself...

3. Originally Posted by isp_of_doom
Alright... I'm not convinced I've got this right as I'm quite new to implicit differentiation myself... but here's what I got step by step (do not take this as correct though).

$\displaystyle \frac{d}{dx}(x^2) + \frac{d}{dx}(2xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(15)$

$\displaystyle 2x + 2y + 2x {\color{red}\frac{dy}{dx}} + \frac{dy}{dx}\frac{d}{dy}(y^2) = 0$

$\displaystyle 2x + 2y + 2x {\color{red}\frac{dy}{dx}} + \frac{dy}{dx}2y = 0$

[snip]
Corrections in red.

But the question becomes trivial if it's observed that the left hand side is $\displaystyle (x + y)^2$ and therefore the equation can be written $\displaystyle x + y = \pm \sqrt{15}$.

4. Sorry Mr. Fantastic, but won't the 2xy alter that $\displaystyle =\pm\sqrt{15}$

5. Originally Posted by isp_of_doom
Sorry Mr. Fantastic, but won't the 2xy alter that $\displaystyle =\pm\sqrt{15}$
$\displaystyle x^2 + 2xy + y^2 = 15 \Rightarrow (x + y)^2 = 15 \Rightarrow ....$

6. so the answer is 1?

7. Originally Posted by Nightasylum