# How do you find #(dy)/(dx)# given #4x^2+4xy=-5x^3y+4#?

##### 1 Answer

# dy/dx = -(15x^2y + 8x + 4y)/(4x + 5x^3) #

#### Explanation:

When we differentiate

However, we only differentiate explicit functions of

Example:

#d/dx(y^2) = d/dy(y^2)dy/dx = 2ydy/dx #

When this is done in situ it is known as implicit differentiation.

Now, we have:

# 4x^2+4xy = -5x^3y+4 #

Implicitly differentiating wrt

# 8x + (4x)(dy/dx) + (4)(y) = (-5x^3)(dy/dx) + (-15x^2)(y) #

# :. 8x + 4xdy/dx + 4y = -5x^3dy/dx -15x^2y #

# :. 4xdy/dx + 5x^3dy/dx = -15x^2y -8x -4y #

# :. (4x + 5x^3)dy/dx = -(15x^2y + 8x + 4y) #

# :. dy/dx = -(15x^2y + 8x + 4y)/(4x + 5x^3) #

**Advanced Calculus**

There is another (often faster) approach using partial derivatives. Suppose we cannot find

# (partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y)) #

So Let

#(partial F)/(partial x) = 8x +4y + 15x^2y #

#(partial F)/(partial y) = 4x + 5x^3 #

And so:

# dy/dx = -(8x +4y + 15x^2y)/(4x + 5x^3) # , as before.