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

Question

Cameron Alexander · Accepted Answer

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

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

We differentiate everything wrt #x#:

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

We can just deal with the #x^3# and the constant term;

# 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 #y# (this is the "Implicit" part of the differentiation

# 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) #

Charles Arocha · Answer

undefined To find $ \frac{dy}{dx} $ by implicit differentiation of $ x^3 - xy + y^2 = 4 $, you differentiate each term with respect to $ x $ and then solve for $ \frac{dy}{dx} $.

Differentiating each term with respect to $ x $, we get:
$$ \frac{d}{dx}(x^3) - \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(4) $$

This simplifies to:
$$ 3x^2 - (x\frac{dy}{dx} + y) + 2y\frac{dy}{dx} = 0 $$

Rearranging terms and solving for $ \frac{dy}{dx} $, we get:
$$ \frac{dy}{dx} = \frac{x - 3x^2 + y}{2y - x} $$