How do you use implicit differentiation to find dy/dx given #x^2y=2#?
1 Answer
Jan 24, 2017
# dy/dx = -(2y)/x #
Explanation:
We have:
#x^2y=2#
Differentiate wrt
# (x^2)(d/dx y) + (d/dx x^2)(y) = d/dx(2) #
# :. x^2dy/dx + 2xy = 0 #
# :. x^2dy/dx = -2xy #
# :. dy/dx = -(2xy)/x^2 #
# :. dy/dx = -(2y)/x #
Strictly speaking we do not need top use implicit differentiation for this equation, as:
# x^2y=2 #
# y=2x^-2 #
Which we can just differentiate to get;
# dy/dx = 2*(-2x^-3) #
# \ \ \ \ \ \ = -2* 2x^-2* x^-1 #
# \ \ \ \ \ \ = -2*y*x^-1 #
# \ \ \ \ \ \ = -(2y)/x # , as above.
