what is the approximate slope of the tangent line to the curve x^(3) +y^(3) =xy at x=1 ?

Question

I'm a little stuck. I believe this involves implicit differentiation. Thanks.

Answer 1

First, solve for y when x=1.

 

1³ + y³ = 1(y)

 

y³ + 1 = y

 

y³ - y + 1 = 0

 

 

Graph the function (y³ - y + 1) on your graphing calculator and find the zero (x-intercept).  This x-intercept will represent the value of y.

 

 

Then, we need to find the derivative of y using implicit differentiation.  Differentiate both sides of the equation using the chain rule.  I will use y' to denote derivative of y.

 

So implicitly deriving,

 

3x² + 3y²y' = y + xy'

 

 

Isolate all the y' terms on one side of the equation all the non-y' terms on the other side.

 

3y²y' - xy' = -3x² + y

 

 

Factor out y'.

 

y'(3y² - x) = -3x² + y

 

y' = (-3x² + y) / (3y² - x)

 

 

Next, plug in x=1 and the value of y (x-intercept you found using graphing calculator) to find y'.