# 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 #