Question

Can anybody help me get the derivative? Question is: Find the tangent line to the folium of Descartes x^3 + y^3 = 6xy at the point (3,3).

I put:

x^3 + y^3 - 6xy = 0

Not sure how to get the derivative of -6xy

Answer 1

product rule

Answer 2

the constant rule allows states that a constant is useless so pull it out and save it for later

Answer 3

the product rule for xy is:

Dx(xy) = x'y + xy' ; where x' = 1 since dx/dx = 1

Answer 4

bring back the -6 to get:

-6(y + xy') = -6y -6xy'

Answer 5

try to see the variable themselves as functions of some other "letter"

x = x(t) and y = y(t)

[x(t) y(t)]' = x'(t) y(t) + x(t) y'(t) ; by the product rule

Answer 6

to wit:

d (x(t)) d(x)
----- = ---
dx dx

x'(t) = dx/dx

x'(t) = 1
............................


d (y(t)) d(y)
----- = ---
dx dx

y'(t) = dy/dx

y'(t) = y'

Answer 7

ok im going to give this a shot. Thank you again. :)