Question

How do you implicit differentiate this? `(xy)^2 + x^3y^2=3y`

Answer 1

Write `(xy)^2` as `x^2y^2`:

`x^2y^2 + x^3y^2=3y`

Differentiate each term with respect to x:

`(d(x^2y^2))/dx+ (d(x^3y^2))/dx=(d(3y))/dx`

The first term requires the use of the product rule:

`(d(x^2y^2))/dx = (d(x^2))/dxy^2+ x^2(d(y^2))/dx`

The first term on the right requires the use of the power rule, `(d(x^n))/dx = nx^(n-1)`:

`(d(x^2y^2))/dx = 2xy^2+ x^2(d(y^2))/dx`

The first term on the right requires the use of the chain rule, `(d(y^2))/dx = (d(y^2))/dydy/dx`:

`(d(x^2y^2))/dx = 2xy^2+ x^2(d(y^2))/dydy/dx`

And the use of the power rule, `(d(y^2))/dy = 2y`:

`(d(x^2y^2))/dx = 2xy^2+ 2x^2y dy/dx`

Return to the equation with the above substitution:

`2xy^2+ 2x^2y dy/dx+ (d(x^3y^2))/dx=(d(3y))/dx`

The next term is very similar except that power of x is 3:

`2xy^2+ 2x^2y dy/dx+ 3x^2y^2+ 2x^3y dy/dx=(d(3y))/dx`

The last term requires that we apply the linear property of differentiation:

`2xy^2+ 2x^2y dy/dx+ 3x^2y^2+ 2x^3y dy/dx=3dy/dx`

Collect all of the terms containing `dy/dx` on the left and all of the other terms on the right:

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

Factor out `dy/dx`:

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

Divide both sides of the equation by the leading factor:

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

Done.