Question

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

Answer 1

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

Explanation:

When we differentiate `y` wrt `x` we get `dy/dx`.

However, we only differentiate explicit functions of `y` wrt `x`. But if we apply the chain rule we can differentiate an implicit function of `y` wrt `y` but we must also multiply the result by `dy/dx`.

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 `x` (applying product rule):

`(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 `y` explicitly as a function of `x`, only implicitly through the equation `F(x, y) = 0` which defines `y` as a function of `x, y = y(x)`. Therefore we can write `F(x, y) = 0` as `F(x, y(x)) = 0`. Differentiating both sides of this, using the partial chain rule gives us

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

So Let `F(x,y) = sin2x^2y^3-3x^3-1`; Then;

`(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.