How do you find #dy/dx# by implicit differentiation of #x^2-y^2=16#?
1 Answer
Nov 23, 2016
Explanation:
If an 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) #
This will typically result in
So for
# d/dx(x^2) - d/dx(y^2) = d/dx(16) #
# :. 2x - dy/dxd/dy(y^2) = 0 #
# :. 2x - 2ydy/dx = 0 #
# :. 2ydy/dx = 2x #
# :. dy/dx = x/y #