what is the approximate slope of the tangent line to the curve x^(3) +y^(3) =xy at x=1 ? I'm a little stuck. I believe this involves implicit differentiation. Thanks
Impicit differentiation it is!
The rule is:
when we differentiate f(y) = y^3, wrt x, then
df/dx = (df/dy)*(dy/dx), i.e.
df/dx = (d(y^3)/dy)*(dy/dx)
df/dx = (3y*2)*(dy/dx)
So, when differentiating f(y) wrt x, simply diffrentiate f(y) wrt y and then multiply that result by y' (=dy/dx)
What we have now is:
x^(3) +y^(3) =xy
differentiating wrt x,
3x^2 + (3y^2).y' = x.y' + y (using the product rule on the lhs there)
3x^2 - y = (x - 3y^2).y'
y' = (3x^2 - y) / (x - 3y^2)
Using again x^(3) +y^(3) = xy.
At x = 1, 1^3 + y^3 = 1.y
y - y^3 = 1
Graphing this cubic and reading off the graph.
At x = 1, y = -1.32 (approx)
Using these values with y' = (3x^2 - y) / (x - 3y^2)
y' = (3.(1^2) - (-1.32)) / (1 - 3.(-1.32)^2)
y' = (3 + 1.32) /( 1 - 5.2272)
y' = 4.32 / (-4.2272)
y' = -1.0219
y = -1.022 (approx)