# Implicit Differentiation

1. Oct 24, 2009

### JBD2

1. The problem statement, all variables and given/known data
Consider:
$$x^3+y^3+2xy=4$$, y=1 when x=1
a.) Find the equation of the tangent line to the curve when x=1.
b.) Find y'' at x=1.
c.) Is the graph of y=f(x) concave up or concave down near x=1?

2. Relevant equations
Any derivative rules...

3. The attempt at a solution
For Part a:

$$y'=\frac{-2y-3x^2}{3y^2+2x}$$

After substituting x=1 and y=1:

$$y'=-1$$

For y=mx+B, B=2

$$y=-x+2$$

For Part b:

$$y''=\frac{vu'-uv'}{v^2}$$

$$u=-2y-3x^2$$
$$u'=-2y'-6x$$
$$v=3y^2+2x$$
$$v'=6yy'+2$$

$$y''=\frac{-(3y^2+2x)(2y'+6x)+(6yy'+2)(2y+3x^2)}{(3y^2+2x)^2}$$

After substituting x=1, y=1, and y'=-1:

$$y''=-1.6$$

For Part c:

Would it be concave down because I just tested y'' with x=1?

I'm mostly unsure about my steps in part b.) and c.), and I want to make sure I'm doing it correctly. Thanks for your help.

2. Oct 24, 2009

### JBD2

Sorry for this unnecessary post, but I don't have much time and need to be sure this is correct.

3. Oct 25, 2009

### HallsofIvy

Staff Emeritus
Yes, that is right. But I wouldn't bother to solve for the general expression for y'.
By implicit differentiation you got 3x2+ 3y2y'+ 2y+ 2xy'= 0. Setting x= y= 1 in that, 3+ 3y+ 2+ 3y'= 5+ 5y'= 0 which gives y'= -1.

Yes, that is correct.

Again, I would work from 3x2+ 3y2y'+ 2y+ 2xy'= 0. Differentiating both sides of that, with respect to x, 6x+ 6y(y')2+ 3yy"+ 2y'+ 2y'+ 2xy"= 0.

Again, setting x= y= 1 and y'= -1,
6(1)+ 6(1)(-1)2+ 3(1)y"+ 4(-1)+ 2(1)y"= 0
6+ 6+ 3y"- 4+ 2y"= 0 gives 5y"= -8, y"= -8/5.

Yes, -8/5= -1.6.

Yes, y"< 0 so the graph is concave down.

4. Oct 25, 2009

### JBD2

Thank you so much, that makes a lot more sense to do it that way. I was wondering though, for your second derivative, is 3yy" supposed to be 3y2y" or is there something I'm forgetting? Thanks again.