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Implicit Differentiation - Tangent Line & Horizontal Tangents

  • Thread starter Pondera
  • Start date
  • #1
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Homework Statement


2181.jpg


Find an equation of the tangent line to this curve at the point (1, -2).

Homework Equations





The Attempt at a Solution



2y' = 3x^2+6x
y' = 3x^2+6x
y'=3/2x^2+3x

y+2=3(x-1)
y+2=3x-3
y=3x-5
 
Last edited:

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
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Hi Pondera

Homework Statement


2181.jpg



Homework Equations





The Attempt at a Solution



2y' = 3x^2+6x
y' = 3x^2+6x
where did the 2 go?
y'=3/2x+3x
where did the power of x go?
y+2=3(x-1)
y+2=3x-3
y=3x-5
I think the arithmetic needs a little work
 
  • #3
13
0
Lane, I divided 3x^2+6x by two and got y'=3/2x^2+3x.

By power of x, I take it you mean power of 2? I neglected to put that in here, but it is on my scratch paper, I appologize.

I don't see where the arithmetic is flawed? I certainly believe that that is likely not the equation/form that I need, but I believe I worked what I figured to be correct out correctly. Can you be more specific?
 
  • #4
lanedance
Homework Helper
3,304
2
ok I don't really understand what you are trying to do, just picked up some misssing parts as discussed

the question is clipped off when I veiw it
 
  • #5
lanedance
Homework Helper
3,304
2
Also the implict derivative w.r.t. x of y2 is not 2y', it will be:

[tex]\frac{d}{dx} y^2 = 2y \frac{dy}{dx} [/tex]

horizontal derivatives where y' = 0
 

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