
#1
Sep2910, 05:03 PM

P: 72

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 x^{3}y^{3}=1 3y^{2}(dy/dx)3x^{2}=0 3y^{2}(dy/dx)=3x2^{2} (dy/dx)=3y^{2}/3x^{2} (dy/dx)=x^{2}/y^{2} slope = (dy/dx) = 1 y(1)= 1(x1) y=x2 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. 



#2
Sep2910, 05:08 PM

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What has x^3y^3=1 got to do with the problem? Your equation is (xy1)^3=x. Find y' using implicit differentiation.




#3
Sep2910, 05:17 PM

P: 72

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 



#4
Sep2910, 05:49 PM

P: 72

Find an equation of the tangent line to the graph of the function f
... i can't figure out how to do the implicit differentiation on this. Could someone help me out with it?




#5
Sep2910, 05:53 PM

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#6
Sep2910, 06:02 PM

P: 72

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 x^{3}y^{3}1^{3} but obviously that's not right since your first response was what did x^{3}y^{3}= 1 have to do anything. So is it 3x^{2}*u and 3y^{2}*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. 



#7
Sep2910, 07:23 PM

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




#8
Sep2910, 08:53 PM

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