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Implicit Differentiation

  • Thread starter Neil6790
  • Start date
  • #1
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Homework Statement


Use implicit differentiation to find the slope of the tangent line to the curve

4x^2-3xy+1y^3=26

at the point (3,2)

The Attempt at a Solution


I attempted the problem and i came up with dy/dx= (-8x+4)/(3y^2) which is wrong.

Need some help with this.
 

Answers and Replies

  • #2
Dick
Science Advisor
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How did you get that?
 
  • #3
HallsofIvy
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Then show us HOW you got that answer!

I suspect you may have messed up a "product rule" but I can't be sure unless you show exactly what you did.
 
  • #4
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This is what i did:

It sounds totally wrong and it looks wrong but i didn't know what to do



(d/dx)(4x^2-3xy+1y^3)=26
(dy/dx)(3y^2)=-8x+4
dy/dx=(-8x+4)/(3y^2)
 
  • #5
djeitnstine
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For your 3xy term you have to use the product rule.... you will find that it will turn out to be -[ (3 dx/dx y') + (3x y') ] product rule being f'g + fg' y' being dy/dx of course
 
  • #6
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For your 3xy term you have to use the product rule.... you will find that it will turn out to be (3 dx/dx y') - (3x y') product rule being f'g + fg' y' being dy/dx of course
I still don't get what you mean. When i differentiate 3xy using the product rule, what should i get? Am i supposed to get (3*(xy)) - (3x*1)? I don't completely get the concept
 
  • #7
djeitnstine
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differentiating 3xy using the product rule (f'g + fg' - in words this is the derivative of f times g plus f times the derivative of g) looks like this [tex]3y\frac{dx}{dx}[/tex] + [tex]3x\frac{dy}{dx}[/tex] Which leaves you with [tex]3y + 3x\frac{dy}{dx}[/tex]
 
  • #8
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Ahhh, I see what you mean now. I did everything, but for the final slope i get -30/21. I have no idea how it's wrong when I did exactly what you told me.
 
  • #9
djeitnstine
Gold Member
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also don't forget that the y^3 differentiates to (3y^2) * (y')
 
  • #10
20
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I was finally able to get the answer which was -6. Thank you very much for the help.
 
  • #11
djeitnstine
Gold Member
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No problem
 

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