Question

How do you implicitly differentiate `-y= x^3y^2-3x^2y^3-7xy^4`?

Answer 1

A useful timesaver is to use `y'` instead of `dy/dx` when differentiating y ... here is how it works:

Explanation:

`-y= x^3y^2-3x^2y^3-7xy^4`

Now, using the chain and product rules ...

`-y'= 3x^2y^2+x^3(2yy')-6xy^3-3x^2(3y^2y')-7y^4-7x(4y^3y')`

Next combine the `y'` terms on the left side of the equation ...

`y'[-1-x^3(2y)+3x^2(3y^2)+7x(4y^3)]=3x^2y^2-6xy^3-7y^4`

Finally, simplify and solve for `y'`

`y'=(3x^2y^2-6xy^3-7y^4)/[9x^2(y^2)+28x(y^3)-x^3(2y)-1]`

hope that helped