How do I find the slope of a tangent line using implicit differentiation?

[Neil6790]

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.

 

[Dick]

How did you get that?

 

[HallsofIvy]

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.

 

[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)

 

[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

 

[Neil6790]

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

 

[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}

 

[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.

 

[djeitnstine]

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

 

[Neil6790]

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

 

[djeitnstine]

No problem