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Chain Rule

If $g(x)$ is Differentiable at the point $x$ and $f(x)$ is Differentiable at the point $g(x)$, then $f\circ g$ is Differentiable at $x$. Furthermore, let $y=f(g(x))$ and $u=g(x)$, then

{dy\over dx}={dy\over du} \cdot {du\over dx}.
\end{displaymath} (1)

There are a number of related results which also go under the name of ``chain rules.'' For example, if $z=f(x,y)$, $x=g(t)$, and $y=h(t)$, then
{dz\over dt}={\partial z\over\partial x}{dx\over dt}+{\partial z\over\partial y}{dy\over dt}.
\end{displaymath} (2)

The ``general'' chain rule applies to two sets of functions
$\displaystyle y_1$ $\textstyle =$ $\displaystyle f_1(u_1, \ldots, u_p)$  
  $\textstyle \vdots$   (3)
$\displaystyle y_m$ $\textstyle =$ $\displaystyle f_m(u_1, \ldots, u_p)$  

$\displaystyle u_1$ $\textstyle =$ $\displaystyle g_1(x_1,\ldots,x_n)$  
  $\textstyle \vdots$   (4)
$\displaystyle u_p$ $\textstyle =$ $\displaystyle g_p(x_1,\ldots,x_n).$  

Defining the $m\times n$ Jacobi Matrix by
\left({\partial y_i\over\partial x_j}\right)=\left[{\matrix{...
...m\over\partial x_n}\cr}}\right],
\hrule width 0pt height 6.8pt
\end{displaymath} (5)

and similarly for $(\partial y_i/\partial u_j)$ and $(\partial u_i/\partial x_j)$ then gives
\left({\partial y_i\over\partial x_j}\right)=\left({\partial...
...rtial u_j}\right)\left({\partial u_i\over\partial x_j}\right).
\end{displaymath} (6)

In differential form, this becomes

dy_1=\left({{\partial y_1\over\partial u_1}{\partial u_1\ove...
...rtial u_p}{\partial u_p\over\partial x_2}}\right)\,dx_2+\ldots
\end{displaymath} (7)

(Kaplan 1984).

See also Derivative, Jacobian, Power Rule, Product Rule


Anton, H. Calculus with Analytic Geometry, 2nd ed. New York: Wiley, p. 165, 1984.

Kaplan, W. ``Derivatives and Differentials of Composite Functions'' and ``The General Chain Rule.'' §2.8 and 2.9 in Advanced Calculus, 3rd ed. Reading, MA: Addison-Wesley, pp. 101-105 and 106-110, 1984.

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© 1996-9 Eric W. Weisstein