Implicit Differentiation

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Homework Statement


Consider:
[tex]x^3+y^3+2xy=4[/tex], 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?

Homework Equations


Any derivative rules...

The Attempt at a Solution


For Part a:

[tex]y'=\frac{-2y-3x^2}{3y^2+2x}[/tex]

After substituting x=1 and y=1:

[tex]y'=-1[/tex]

For y=mx+B, B=2

[tex]y=-x+2[/tex]

For Part b:

[tex]y''=\frac{vu'-uv'}{v^2}[/tex]

[tex]u=-2y-3x^2[/tex]
[tex]u'=-2y'-6x[/tex]
[tex]v=3y^2+2x[/tex]
[tex]v'=6yy'+2[/tex]

[tex]y''=\frac{-(3y^2+2x)(2y'+6x)+(6yy'+2)(2y+3x^2)}{(3y^2+2x)^2}[/tex]

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

[tex]y''=-1.6[/tex]

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.
 

Answers and Replies

  • #2
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Sorry for this unnecessary post, but I don't have much time and need to be sure this is correct.
 
  • #3
HallsofIvy
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Homework Statement


Consider:
[tex]x^3+y^3+2xy=4[/tex], 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?

Homework Equations


Any derivative rules...

The Attempt at a Solution


For Part a:

[tex]y'=\frac{-2y-3x^2}{3y^2+2x}[/tex]

After substituting x=1 and y=1:

[tex]y'=-1[/tex]
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.

For y=mx+B, B=2

[tex]y=-x+2[/tex]
Yes, that is correct.

For Part b:

[tex]y''=\frac{vu'-uv'}{v^2}[/tex]

[tex]u=-2y-3x^2[/tex]
[tex]u'=-2y'-6x[/tex]
[tex]v=3y^2+2x[/tex]
[tex]v'=6yy'+2[/tex]

[tex]y''=\frac{-(3y^2+2x)(2y'+6x)+(6yy'+2)(2y+3x^2)}{(3y^2+2x)^2}[/tex]
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.

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

[tex]y''=-1.6[/tex]
Yes, -8/5= -1.6.

For Part c:

Would it be concave down because I just tested y'' with x=1?
Yes, y"< 0 so the graph is concave down.

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.
 
  • #4
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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.

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.
 

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