# How do you use implicit differentiation to find #(dy)/(dx)# given #sin2x^2y^3=3x^3+1#?

##### 1 Answer

# dy/dx = (9x^2 - 4xcos(2x^2)y^3)/(3sin(2x^2)y^2) #

#### 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:

# sin2x^2y^3=3x^3+1 #

Implicitly differentiating wrt

# (sin2x^2)(d/dx y^3) + (d/dx sin2x^2)(y^3) = 9x^2#

# :. (sin2x^2)(3y^2dy/dx ) + (4xcos2x^2)(y^3) = 9x^2#

# :. 3sin(2x^2)y^2dy/dx + 4xcos(2x^2)y^3 = 9x^2#

# :. 3sin(2x^2)y^2dy/dx = 9x^2 - 4xcos(2x^2)y^3 #

# :. dy/dx = (9x^2 - 4xcos(2x^2)y^3)/(3sin(2x^2)y^2) #

**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) = 4xcos(2x^2)y^3-9x^2 #

#(partial F)/(partial y) = sin(2x^2)(3y^2) #

And so:

# dy/dx = -(4xcos(2x^2)y^3-9x^2)/(sin(2x^2)(3y^2)) #

# " " = (9x^2-4xcos(2x^2)y^3)/(3sin(2x^2)y^2) # , as before.