# Proof of Multivariable chain rule

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

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

HallsofIvy
Homework Helper
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.

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.

lurflurf
Homework Helper
Same proof as the single variable chain rule.

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.

Makes sense. thanks.