# The Chain Rule

Prequisites: Partial Derivatives

Back in basic calculus, we learned how to use the chain rule on single variable functions. Now we want to be able to use the chain rule on multi-variable functions.

Lets start with the two-variable function and then generalize from there. Let *z*=*f*(*x,y*) be our two-variable function. Lets say that both *x* and *y* can be expressed as functions of another variable *t*. *x*=*g*(*t* and *y*=*h*(*t*. Then *z* is a function of *t*, say *z*=*f*[*g*(*t*),*h*(*t*)] = *F*(*t*). Then the formula for the derivative of *z* with respect to *t* is

Don't let the fact that I switched notation throw you

### Proof

Lets first notice a few things...
Now, by linear approximation we know
where the above error functions go to zero as **delta** x and **delta** y go to zero. We divide by **delta** t and get,
##

Now as we let **delta** t go to zero, **delta** x, **delta** y, and the error functions go to zero and we get the expression.
This formula extends as one would expect. If *z*=*f*(*x*_{1},*x*_{2},...x_{n}) and *x*_{1} through *x*_{n} can be expressed as a function of a variable *t*, then,
It is not necessary that *x*_{1} through *x*_{n} be only functions of *t*. They can be functions of other variables, too. Say *x*_{1} through *x*_{n} are functions of two variables *t* and *u*. Then the partial derivatives of *z* are given by
These rules define the **chain rules
**

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Last modified 8 August 1997