Implicit Differentiation

1. Oct 17, 2008

Neil6790

1. The problem statement, all variables and given/known data
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)

3. 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.

2. Oct 17, 2008

Dick

How did you get that?

3. Oct 17, 2008

HallsofIvy

Staff Emeritus
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. Oct 17, 2008

Neil6790

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. Oct 17, 2008

djeitnstine

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. Oct 17, 2008

Neil6790

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. Oct 17, 2008

djeitnstine

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 $$3y\frac{dx}{dx}$$ + $$3x\frac{dy}{dx}$$ Which leaves you with $$3y + 3x\frac{dy}{dx}$$

8. Oct 17, 2008

Neil6790

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. Oct 17, 2008

djeitnstine

also don't forget that the y^3 differentiates to (3y^2) * (y')

10. Oct 17, 2008

Neil6790

I was finally able to get the answer which was -6. Thank you very much for the help.

11. Oct 17, 2008

No problem