# How do you find #dy/dx# by implicit differentiation of #x^3-xy+y^2=4#?

##### 1 Answer

Nov 28, 2016

#### Explanation:

# x^3-xy+y^2=4 #

We differentiate everything wrt

# d/dx(x^3) - d/dx(xy) + d/dx(y^2) = d/dx(4) #

We can just deal with the

# 3x^2 - d/dx(xy) + d/dx(y^2) = 0 #

For the second term we apply the product rule;

# 3x^2 - { (x)(d/dx(y)) +(d/dx(x))(y } + d/dx(y^2) = 0 #

# :. 3x^2 - { xdy/dx +(1)(y) } + d/dx(y^2) = 0 #

# :. 3x^2 - xdy/dx + y + d/dx(y^2) = 0 #

And for the last term we use the chain rule so that we can differentiate wrt

# 3x^2 - xdy/dx + y + dy/dxd/dy(y^2) = 0 #

# :. 3x^2 - xdy/dx + y + dy/dx2y = 0 #

# :. xdy/dx - 2ydy/dx= 3x^2 + y #

# :. (x-3y)dy/dx = 3x^2 + y #

# :. dy/dx = (3x^2 + y)/(x-3y) #