# Thread: implicit differentiation f(x,y,z)

1. ## implicit differentiation f(x,y,z)

Let y be a function of x satisfying $F(x,y, x+y)=0$ where $F(x,y,z)$ is a given function. find a formula for $\frac{dy}{dx}$

$F_{x} = (1, 0, 1), F_{y} = (0,1,1)$

2. ## Re: implicit differentiation f(x,y,z)

I know $\frac{dy}{dx} = \frac{-f_{x}}{{f_y}}$

but how does that change with 3 variables

3. ## Re: implicit differentiation f(x,y,z)

$\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}$

$\frac{-\frac{\partial f}{\partial x}(1+1)}{\frac{\partial f}{\partial y}(1+1)} = \frac{dy}{dx}$

or

$\frac{-\frac{\partial f}{\partial x}\frac{\partial f}{\partial z}}{\frac{\partial f}{\partial y}} = \frac{dy}{dx}$

??

4. ## Re: implicit differentiation f(x,y,z)

If x, y, and z were independent variables, then z would be treated like a constant while calculating dy/dx.

But if F(z, y, z)= F(x, y, x+ y)= 0, then z= x+y, and, differentiating with respect to x, $\frac{\partial F}{\partial x}+ \frac{\partial F}{\partial y}\frac{dy}{dx}+ \frac{\partial F}{\partial z}\left(\frac{dy}{dx}+ 1\right)= 0$.