Prob Expert: Chain rule for several variables

The chain rule works for functions of more than one variable. Consider the function $z = f (x, y)$ where $x = g (t)$ and $y = h (t)$ , then

${\partial z \over \partial t}={\partial f \over \partial x}{dx \over dt}+{\partial f \over \partial y}{dy \over dt}$

Suppose that each function of $z = f (u, v)$ is a two-variable function such that $u = h (x, y)$ and $v = g (x, y)$ , and that these functions are all differentiable. Then the chain rule would look like:

${\partial z \over \partial x}={\partial z \over \partial u}{\partial u \over \partial x}+{\partial z \over \partial v}{\partial v \over \partial x}$

${\partial z \over \partial y}={\partial z \over \partial u}{\partial u \over \partial y}+{\partial z \over \partial v}{\partial v \over \partial y}$

If we considered $\vec r = (u,v)$ above as a vector function, we can use vector notation to write the above equivalently as the dot product of the gradient of f and a derivative of $\vec r$ :

$\frac{\partial f}{\partial x}=\vec \nabla f \cdot \frac{\partial \vec r}{\partial x}$

More generally, for functions of vectors to vectors, the chain rule says that the Jacobian matrix of a composite function is the product of the Jacobian matrices of the two functions:

$\frac{\partial(z_1,\ldots,z_m)}{\partial(x_1,\ldots,x_p)} = \frac{\partial(z_1,\ldots,z_m)}{\partial(y_1,\ldots,y_n)} \frac{\partial(y_1,\ldots,y_n)}{\partial(x_1,\ldots,x_p)}$

Prob Expert

Wednesday, February 28, 2007

Chain rule for several variables

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