The Chain Rule
You probably remember the derivatives of
, and
. But what about functions like 
,
, or 
? How do we take the derivative
of compositions of functions?
The Chain Rule allows us to use our knowledge of the derivatives of functions f(x) and g(x) to find the derivative of the composition f(g(x)):
![\fbox{\begin{minipage}{4in}
\par
Suppose a function $g(x)$\ is differentiable at...
...eft[ f(u) \right] & = & f'(u)\frac{du}{dx}.
\end{eqnarray*}\par
\end{minipage}}](img9.png)
The three formulations of the Chain Rule given here are identical in meaning. In words, the derivative of f(g(x)) is the derivative of f, evaluated at g(x), multiplied by the derivative of g(x).
, we identify 
. Then
![\begin{eqnarray*}
\frac{d}{dx}\left[ \sin(2x-1)\right] & = & \frac{d}{dx} \left[...
...1 \right] \\
& = & \cos (u) \cdot 2 \\
& = & 2 \cos (2x-1).
\end{eqnarray*}](img12.png) |
, we
identify 
. Then
![\begin{eqnarray*}
\frac{d}{dx} \left[\left( 3x^{2} - 4x + 1 \right)^{8}\right] &...
...ot (6x-4) \\
& = & 8 (6x-4)\left( 3x^{2} - 4x + 1 \right)^{7}.
\end{eqnarray*}](img16.png) |
, we identify 
. Then
![\begin{eqnarray*}
\frac{d}{dx} \left[ e^{-x^{2}} \right] & = & \frac{d}{du}\left...
...t[ -x^2 \right] \\
& = & e^u \cdot (-2x) \\
& = & -2xe^{-x^2}.
\end{eqnarray*}](img19.png) |
Sometimes you will need to apply the Chain Rule several times in order to differentiate a function.
.
Let g(x) be differentiable at x and f(x) be differentiable at
f(g(x)). Then, if y = f(g(x)) and u = g(x),
