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The Chain Rule
You probably remember the derivatives of $\sin (x)$, $x^{8}$, and
$e^{x}$. But what about functions like $\sin (2x-1)$,
$(3x^{2}-4x+1)^{8}$, or $e^{-x^{2}}$? 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 $f(g(x))$:
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Suppose a function $g(x)$ is differentiable at $x$ and $f(x)$ is
differentiable at $g(x)$. Then the composition $f(g(x))$ is
differentiable at $x$.
Letting $y=f(g(x))$ and $u=g(x)$,
$$
\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}.
$$
Using alternative notation,
\begin{eqnarray*}
\frac{d}{dx}\left[ f(g(x)) \right] & = & f'(g(x))g'(x), \\
\frac{d}{dx}\left[ f(u) \right] & = & f'(u)\frac{du}{dx}.
\end{eqnarray*}
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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 $g(x)$.
Examples
- To differentiate $\sin (2x-1)$, we identify $u=2x-1$. Then
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\begin{eqnarray*}
\frac{d}{dx}\left[ \sin(2x-1)\right] & = & \frac{d}{du} \left[ \sin
(u) \right] \cdot \frac{d}{dx} \left[ 2x-1 \right] \\
& = & \cos (u) \cdot 2 \\
& = & 2 \cos (2x-1).
\end{eqnarray*}
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\begin{eqnarray*}
f(x) & = & \sin (x) \\
g(x) & = & 2x-1 \\
f(g(x)) & = & \sin (2x-1)
\end{eqnarray*}
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- To differentiate $\left( 3x^{2} - 4x + 1 \right)^{8}$, we
identify $u=3x^2-4x+1$. Then
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\begin{eqnarray*}
\frac{d}{dx} \left[\left( 3x^{2} - 4x + 1 \right)^{8}\right] & = &
\frac{d}{du}\left[ u^8 \right] \cdot \frac{d}{dx}\left[ 3x^2-4x+1
\right] \\
& = & 8u^7 \cdot (6x-4) \\
& = & 8 (6x-4)\left( 3x^{2} - 4x + 1 \right)^{7}.
\end{eqnarray*}
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\begin{eqnarray*}
f(x) & = & x^8 \\
g(x) & = & 3x^{2} - 4x + 1 \\
f(g(x)) & = & \left( 3x^{2} - 4x + 1 \right)^{8}
\end{eqnarray*}
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- To differentiate $e^{-x^{2}}$, we identify $u=-x^{2}$. Then
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\begin{eqnarray*}
\frac{d}{dx} \left[ e^{-x^{2}} \right] & = & \frac{d}{du}\left[ e^u
\right] \cdot \frac{d}{dx}\left[ -x^2 \right] \\
& = & e^u \cdot (-2x) \\
& = & -2xe^{-x^2}.
\end{eqnarray*}
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\begin{eqnarray*}
f(x) & = & e^x \\
g(x) & = & -x^2 \\
f(g(x)) & = & e^{-x^{2}}
\end{eqnarray*}
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Sometimes you will need to apply the Chain Rule several times in order
to differentiate a function.
Example
We will differentiate $\sqrt{\sin^{2} (3x) + x}$.
$$
\begin{array}{rclc}
\frac{d}{dx}\left[ \sqrt{\sin^{2} (3x) + x}\right] & = &
\frac{1}{2\sqrt{\sin^{2} (3x) + x}} \cdot \frac{d}{dx}\left[ \sin^{2} (3x) + x
\right] & f(u) = \sqrt{u}\\
& = & \frac{1}{2\sqrt{\sin^{2} (3x) + x}} \cdot \left( 2 \sin (3x)
\frac{d}{dx} \left[ \sin (3x) \right] + 1 \right) &
\begin{array}{rcl}
f(u) & = & u^2 \\
\frac{d}{dx} [x] & = & 1
\end{array} \\
& = & \frac{1}{2\sqrt{\sin^{2} (3x) + x}} \cdot \left( 2 \sin (3x)
\cos (3x) \frac{d}{dx} [3x] + 1 \right) & f(u) = \sin (u) \\
& = & \frac{1}{2\sqrt{\sin^{2} (3x) + x}} \cdot \left( 2 \sin (3x)
\cos (3x) \cdot 3 + 1 \right) \\
& = & \displaystyle\frac{6 \sin (3x) \cos (3x) + 1}{2\sqrt{\sin^{2} (3x) + x}}
\end{array}
$$
Key Concepts
Let $g(x)$ be differentiable at $x$ and $f(x)$ be differentiable at
$f(g(x))$. Then, if $y=f(g(x))$ and $u=g(x)$,
$$
\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}.
$$
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