# Implicit Differentiation

1. Sep 5, 2009

### manenbu

1. The problem statement, all variables and given/known data

Assume that the following equation define the implicit function y=(x). Find the its derivative:
x2 + 2xy - y2 = a2
y'=?
y''=?

2. Relevant equations

$$\frac{dy}{dx} = -\frac{F_x}{F_y}$$

3. The attempt at a solution

so for the first derivative I express that equation as F = x2 + 2xy - y2 - a2 = 0 and using the rule from above I get:
$$y^{'} = -\frac{x+y}{x-y}$$ which is correct.
For the second derivative the answer should be:
$$y^{''} = \frac{2a^{2}}{(x-y)^{3}}$$
But I don't understand how to get there. Where did the 2a2 come from? the 3 hints to me that I need to make a derivative of the fraction, but I can't seem to get anything useful.

2. Sep 5, 2009

### latentcorpse

you can just go through and differentiate everything w.r.t. x

$x^2+2xy-y^2=a^2$

so, remembering to use the product rule on the second term and the chain rule on the third,

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

then collect $\frac{dy}{dx}$ terms as follows:

$2(x+y)=2(y-x) \frac{dy}{dx} \Rightarrow \frac{dy}{dx}=y'=\frac{x+y}{y-x}$ which is what you get.

just do the same again to get the second derivative.

3. Sep 5, 2009

### manenbu

Ok got it right this time.
Just a few questions - why do you add the differential to the y and not the x?
And in the chain rule - where is it?
Am I correct thinking that the chain rule should be (dy/dx)(dx/dx) which cancels to dy/dx?

4. Sep 5, 2009

### Hootenanny

Staff Emeritus
All latentcorpse did, was differentiate the whole expression with respect to x. Explicitly,

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

$$\frac{d}{dx}x^2 + 2\frac{d}{dx}\left(xy\right) - \frac{d}{dx}y^2 = 0$$

Using the product rule on the second term and the chain rule on the final term,

$$2x + 2y\frac{d}{dx}x + 2x\frac{dy}{dx} - \frac{d}{dy}y^2\cdot\frac{dy}{dx} = 0$$

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

Do you follow?

5. Sep 5, 2009

### manenbu

Ok - everything is understood now. Thanks!