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Chain rule

By definition,

\begin{displaymath}
\frac{\mathrm{d}}{\mathrm{d}x} f(g(x)) = \lim_{h \to 0} \frac{f(g(x+h)) -
f(g(x))}{h}
\end{displaymath}

Upon substituting $a = x + h$, This can be written as

\begin{eqnarray*}
\lim_{x \to a} \frac{f(g(x)) - f(g(a))}{x - a} & = & \lim_{x \...
...lim_{x \to a} \frac{g(x) - g(a)}{x - a} \\
& = & f'(g(x)) g'(x)
\end{eqnarray*}



(N.B. This is only a sketch proof.) This can also be written

\begin{displaymath}
\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t}
\frac{\mathrm{d}t}{\mathrm{d}x}
\end{displaymath}

This result is used to differentiate functions of functions, such as \( y = \sin{(x^2 + 2)} \). In this case, we put \( y = \sin{u}, u = x^2 +
2 \), so that \( \frac{\mathrm{d}y}{\mathrm{d}u} = \cos{u}, \frac{\mathrm{d}u}{\mathrm{d}x} = 2x \). Hence, \( \frac{\mathrm{d}y}{\mathrm{d}x} = 2x \cos{u} = 2x \cos{(x^2 + 2)} \).


Alexander Frolkin 2001-03-13