# Implicit differentiation

1. Dec 6, 2005

$$x^2+y^2+r^2-2s=13=0$$
$$x^3-y^3-r^3+3s+59=0$$

How do I find the partial derivatives of x(r,s) or y(r,s) implicitly? I tried implicit differentiation and I got 2 different answers for either. Can someone show me any of the 4 derivatives step-by-step?

2. Dec 6, 2005

### marlon

Please, show us what you have done

marlon

3. Dec 6, 2005

$$2x\frac{\partial x}{\partial r}+2r=0$$
$$3x^2\frac{\partial x}{\partial r}-3r^2=0$$

$$2x\frac{\partial x}{\partial r}+2r=3x^2\frac{\partial x}{\partial r}-3r^2$$

$$\frac{2r+3r^2}{3x^2-2x}=\frac{\partial x}{\partial r}$$

This is what I get if I equate them. If I solve for dx/dr individually and then add the equations, I get something else.

4. Dec 6, 2005

An answer before 2PM tomorrow will be appreciated! (Final exam)

5. Dec 6, 2005

### NateTG

$$F(x,y,s,r)=x^2+y^2+r^2-2s+13=0[/itex] So [tex]\frac{\partial x}{\partial s}=-\frac{\frac{\partial F}{\partial s}}{\frac{ \partial F}{\partial x}}=-\frac{-2}{2x}=\frac{1}{x}$$
so
$$\frac{\partial}{\partial r}\frac{\partial x}{\partial s}=-\frac{1}{x^2} \times \frac{\partial x}{\partial r}$$
And
$$\frac{\partial x}{\partial r}=-\frac{2r}{2x}=-\frac{r}{x}$$
so we have
$$\frac{\partial}{\partial r \partial s} =\frac{r}{x^3}$$

6. Dec 6, 2005

### HallsofIvy

Staff Emeritus
y is also a function of r. You are differentiating both sides of the equation with respect to r. The derivative of x2+ y2+ r2- 2s+ 13= 0 with respect to r is
$$2x\frac{\partial x}{\partial r}+ 2y\frac{\partial y}{\partial x}+ 2r= 0$$

7. Dec 6, 2005

I think it should be dy/dr. It gives me the right answer when it is. Thanks everyone!

Last edited: Dec 6, 2005
8. Dec 7, 2005

### NateTG

Potentially stupid question, but that doesn't make sense to me. I would expect that to be:
$$2x\frac{\partial x}{\partial r}+ 2y\frac{\partial y}{\partial r} + 2r - 2 \frac{\partial s}{\partial r}=0$$

Since $s$ is also a function of $r$. Can you clarify why there is no $\frac{\partial s}{\partial x}$ term?