How do you implicitly differentiate #7=1-(y-x)/y^2#?
1 Answer
Nov 12, 2016
# dy/dx = 1/(1+12y) #
Explanation:
The equation does not express
# g(y) = f(x) #
# :. d/dx g(y) = d/dx f(x) #
# :. dy/dx d/dy g(y) = f'(x) #
# :. g'(y) dy/dx= f'(x) #
So for
# -6 = (y-x)/y^2#
# :. -6y^2 = y-x #
# :. -12ydy/dx = dy/dx-1 #
# :. dy/dx+12ydy/dx = 1 #
# :. (1+12y)dy/dx = 1 #
# :. dy/dx = 1/(1+12y) #