1 Answer | Add Yours

Top Answer

beckden's profile pic

beckden | High School Teacher | (Level 1) Educator

Posted on

You want to find the derivative?

We first use the product property


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

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

Now what confuses most students is what is `(dy^2)/(dx)`

We use the chain rule to get for example `(du^2)/(dx) = 2u (du)/(dx)` correct, so
`(dy^2)/(dx) = 2y (dy)/(dx)` so the left side of our equation is

`x^2(dy)/(dx) + 2xy + 2xy(dy)/(dx) + y^2` since `(dx)/(dx)=1`

The right side of our equation is `(d(6))/(dx) = 0`

So our equation differentiated is

`x^2(dy)/(dx) + 2xy + 2xy(dy)/(dx) + y^2 = 0`

No we solve for `(dy)/(dx)`

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

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

`(dy)/(dx) = -(y^2 + 2xy)/(x^2 + 2xy)` which is our final answer

We’ve answered 319,807 questions. We can answer yours, too.

Ask a question