# Math Help - Implicit differentiation. xy=1.

1. ## Implicit differentiation. xy=1.

I'm taking a quote from my textbook:

"We can think of the equation xy=1 as stating that two functions of x (namely, xy and 1) are equal. It follows that the derivatives of these functions are equal, so

$x\frac{dy}{dx}+y=0$"

This is stated very plainly and I am finding that I'm not following what is being said here. Maybe its the notation that I don't understand. I'm trying to figure out how xy=1 transformed into the equation set to zero. For some reason y is left as y. I'm seeing sometimes that y gets changed to $\frac{dy}{dx}$. I'm trying to follow my textbook the best I can. Can someone explain this to me?

2. ## Re: Implicit differentiation. xy=1.

Hey sepoto.

The basic idea is to use the product and the chain rule. We have as follows:

d/dx [xy] = dx/dx*y + x*dy/dx = d/dx (1) = 0 by the product rule d/dx [uv] = du/dx*v + u*dv/dx. This gives us

y + x*dy/dx = 0.

Remember the chain rule that d/dx u(v(x)) = dv/dx * du(v(x))/dx.

3. ## Re: Implicit differentiation. xy=1.

How is $\frac{dx}{dx}*y$ change to just y?

4. ## Re: Implicit differentiation. xy=1.

The derivative of $x$ with respect to $x$ is 1, and $1\cdot y = y$.

5. ## Re: Implicit differentiation. xy=1.

And, of course, the equation is "set equal to 0" because the derivative of the constant, 1, on the right, is 0.