Here are some chain rule examples. In these examples, I’ll already use the rule \(\frac{d}{dx} \ln x =
\frac{1}{x}\text{,}\) even though the rule will be proved later in this section. In the first example, the outside function is \(e^u\) and the inside function is \(ax\text{.}\)
\begin{equation*}
\frac{d}{dx} e^{ax} = \frac{d}{du} e^u \Bigg|_{u = ax}
\frac{d}{dx} (ax) = ae^{ax}
\end{equation*}
In the next, the outside function is \(\sin u\) and the inside functon is \(3x\text{.}\) After the derivatives, I’ll move the constant to the front of the function. This isn’t necessary, but fits the general conventions of writing constants and polynomials to the left of trigonometric or exponential functions.
\begin{equation*}
\frac{d}{dx} \sin 3x = \frac{d}{du} \sin u \Bigg|_{u =
3x} \frac{d}{dx} (3x) = 3 \cos 3x
\end{equation*}
In the next, the outside function is \(e^u\) and the inside function is \(x^2 + 2\text{.}\)
\begin{equation*}
\frac{d}{dx} e^{x^2+2} = \frac{d}{du} e^u \Bigg|_{u =
x^2 + 2} \frac{d}{dx} (x^2 + 2) = (2x) e^{x^2+2}
\end{equation*}
In the next, the outside function is \(\ln u\) and the inside function is \(5x^3 + 7x + 1\text{.}\)
\begin{equation*}
\frac{d}{dx} \ln (5x^3 + 7x + 1) = \frac{d}{du} \ln u
\Bigg|_{u = 5x^3 + 7x + 1} \frac{d}{dx} (5x^3 + 7x + 1) =
\frac{15x^2 + 7}{5x^3 + 7x + 1}
\end{equation*}
In the last example, this is a triple compsition. I use the chain rule twice, making the outside function as easy as possible in each step. The first outside function is \(\cos
u\) and the second outside function is \(e^v\text{.}\)
\begin{align*}
\frac{d}{dx} \cos ( e^{12 x}) \amp = \frac{d}{du} \cos
y \Bigg|_{u=e^{12 x}} \frac{d}{dx} e^{12 x} = -\sin u
\Bigg|_{u=e^{12x}} \\
\amp = -\sin (e^{12 x}) \frac{d}{dv} e^v \Bigg|_{v = 12
x} \frac{d}{dv} 12 x = -\sin (e^{12 x}) e^v \Bigg|_{v =
12x} (12) \\
\amp = -12 \sin (e^{12 x}) e^{12 x}
\end{align*}
