Find an equation of the tangent line to the graph of the function f

A_Munk3y

1. The problem statement, all variables and given/known data
Find an equation of the tangent line to the graph of the function f defined by the following equation at the indicated point.
(x - y - 1)³ = x; (1, -1)



3. The attempt at a solution
x³-y³=1
3y²(dy/dx)-3x²=0
3y²(dy/dx)=3x2²
(dy/dx)=3y²/3x²
(dy/dx)=x²/y²
slope = (dy/dx) = 1

y-(-1)= 1(x-1)
y=x-2

thats wrong -.-
i tried making it dy/dx = x/y and still wrong.
where am i messing up? I'm pretty sure it has to do with the x at the end but i have no idea what to do with it!
We just learned derivatives so I'm still messing up with them.

Dick

What has x^3-y^3=1 got to do with the problem? Your equation is (x-y-1)^3=x. Find y' using implicit differentiation.

A_Munk3y

Yea, i thought i was deriving it wrong :) Ok, let me go back and look at my notes to see how to do implicit differentiation.
Thanks

A_Munk3y

... i can't figure out how to do the implicit differentiation on this. Could someone help me out with it?

Dick

The derivative of u^3 is 3*u^2*u'. Put u=x-y-1. What do you get? Is that the part that's confusing you?

A_Munk3y

I'm getting confused on the dx/dy part. I'm not even sure i understand how to do this right.
I thought i had to have them all cubed then get the derivative of x³-y³-1³ but obviously that's not right since your first response was what did x³-y³= 1 have to do anything.

So is it 3x²*u and 3y²*u -1? and then i add the dx/dy part somewhere? Or am i just way off here :(
Sorry, we just learned this stuff today and it still hasn't really sunk in.

Char. Limit

Why would you implicitly differentiate? Since 3 is an odd power, you can solve for y without losing any information...

Dick

Good point. If you aren't comfortable with implicit differentiation, try it that way. They both work.