# Implicit Differentiation

1. Jan 8, 2005

Hello all

Given: x^2 + xy + y^2 - 7 = 0, solve for y using the quadratic forumula. Then find dy/dx at P(1,2) from a function of the form f(x).

My solution:

y = -x (+/-) sqrt( x^2 - 28) / 2.

I am not sure if this is correct. After solving for y, do you have to implicitly take the derivative of y?

2. Jan 8, 2005

### marlon

you have to calculate the derivative with repsect to x of this given function. So just look at y as some function dependent on x ; so you have dy/dx...and dy²/dx = 2ydy/dx and so on...

2x + y + xdy/dx + 2ydy/dx = 0

you get this too ???

regards
marlon

3. Jan 8, 2005

### dextercioby

$$y^{2}+xy+x^{2}-7=0$$
Use the quadratic formula to find:
$$y_{1,2}(x) =\frac{-x\pm\sqrt{28-3x^{2}}}{2}$$
(1)
The condition that the point (1,2) should be on the graph of "y" yields:
$$y_{1}(1)=\frac{-1+\sqrt{28-3}}{2}=\frac{-1+5}{2}=2$$(2)

$$y_{2}(1)=\frac{-1-5}{2}=-3$$ (3)

So you need to chose the "+" sign from the explicitation.
$$y(x)=\frac{1}{2}(\sqrt{28-3x^{2}}-x)$$

Compute its derivative and make "x=1" in the result.

Daniel.

Last edited: Jan 8, 2005
4. Jan 8, 2005

### marlon

GUYS GUYS what are you doing...

Your solution is NOT an implicit derivation. You don't need the quadratic formula at all. Besides the motivation, dextercioby for taken the + value is not correct.

What you need to do is look at y as y(x) and calculate the derivative of the given formula with respect to x. This is the IMPLICIT part...the derivavtive that is asked is in a formula itself

x²+xy+y²-7=0
derivative with respect to x yields

2x + y + x(dy/dx) + 2y(dy/dx) = 0

or

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

Then fill in the given x and y values in the right hand side and all is done...

regards
marlon

5. Jan 8, 2005

how did you get 3x^2 for the term in the quadratic forumula dextercioby?

Thanks

6. Jan 8, 2005

### dextercioby

The discriminant is $\Delta=b^{2}-4ac [/tex] Compute it,with [itex] a=1;b=x;c=x^{2}-7$

Daniel.

7. Jan 8, 2005