Implicit differentiation.

**Nightasylum** · October 3rd 2009, 11:32 AM

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

**isp_of_doom** · October 3rd 2009, 11:46 PM

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).

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

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

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

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

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

**mr fantastic** · October 3rd 2009, 11:53 PM

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).

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

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

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

[snip]

Corrections in red.

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

**isp_of_doom** · October 4th 2009, 12:45 AM

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

**mr fantastic** · October 4th 2009, 12:56 AM

Originally Posted by isp_of_doom

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

$x^2 + 2xy + y^2 = 15 \Rightarrow (x + y)^2 = 15 \Rightarrow ....$

**Nightasylum** · October 4th 2009, 10:38 AM

so the answer is 1?

**mr fantastic** · October 4th 2009, 06:29 PM

Originally Posted by Nightasylum

so the answer is 1?

No. Try again and be more careful.

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