# Proof of Multivariable chain rule

1. Nov 7, 2011

### dimension10

I was wondering how to prove the multivariable chain rule

$$\frac{\mbox{d}z}{\mbox{d}t}=\frac{\partial z}{\partial y}\frac{\mbox{d}y}{\mbox{d}t}+\frac{\partial z}{\partial x}\frac{\mbox{d}x}{\mbox{d}t}$$

where $$z=z(x(t),y(t))$$

I don't really need an extremely rigorous proof, but a slightly intuitive proof would do.

Also how does one prove that if z is continuous, then

$$\frac{{\partial}^{2}z}{\partial x \partial y}=\frac{{\partial}^{2}z}{\partial y \partial x}$$

2. Nov 7, 2011

### HallsofIvy

As for your second question, one doesn't- what you have written is not true. If z is only continuous, the partial derivative, much less the second derivatives, may not even exist. What you need is that the second derivatives are continuous.

3. Nov 7, 2011

### JHamm

The MIT OCW videos on multivariable calculus have video which covers this: http://ocw.mit.edu/courses/mathemat...fall-2007/video-lectures/lecture-12-gradient/

To summarize the argument (though I doubt this is particularly rigorous)
$$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz$$
Then, if we assume that f, x, y and z are all functions of t we divide by dt
$$\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}$$
Again this is not rigorous, if you want an idea of why this could be true think of the way a vector is composed by summing orthogonal parts; take the amount f will change when you move an amount along the x direction, multiplied by the amount x will move when you change t a certain amount, then repeat this for y and z and sum for the final change in f.

4. Nov 7, 2011

### lurflurf

Same proof as the single variable chain rule.

5. Nov 8, 2011

### dimension10

Makes sense. thanks.