On this screen, we’re going to use the Chain Rule to find yet more derivatives. This will both (1) give you even more crucial practice at using the Chain Rule, and (2) will let you find the derivative of things like $2^x,$ $\sec x,$ $\cot x,$ and so forth.

Let’s see first how we can easily extend our new knowledge that $\dfrac{d}{dx}e^u = e^u \dfrac{du}{dx}$ to find the derivative of $a^x,$ now much more easily than we could before we had the Chain Rule as a tool.

Practice Problem #1: Derivative of \(a^x\)

Show that $\dfrac{d}{dx}a^x = a^x \ln a$, where $a$ is a constant and $a > 0$. (For example, $\dfrac{d}{dx}2^x = 2^x \ln 2.$)

Hint: Start with $a^x = e^{\ln a^x}$, and use $\dfrac{d}{dx}e^u = e^u \cdot \dfrac{du}{dx}.$

Show/Hide Solution

Start from the hint, and recall that $\ln a^x = x \ln a$:
\begin{align*}
\frac{d}{dx}a^x &= \frac{d}{dx}\left(e^{\ln a^x}\right) &&\text{[Recall that $\ln a^x = x\ln a$]} \\ \\
&= \frac{d}{dx}\left(e^{x \ln a} \right) &&\left[\text{Recall } \dfrac{de^u}{dx} = e^u \dfrac{du}{dx}. \text{Here } u = x \ln a.\right]\\ \\
&= e^{x \ln a} \cdot \left(\frac{d}{dx} (x \ln a) \right) &&\text{[Remember that $a$ is a constant, so $\ln a$ is also a constant.]}\\ \\
&= e^{x \ln a} \cdot \ln a \left(\frac{d}{dx} (x) \right) \\ \\
&= e^{x \ln a} \cdot \ln a\\ \\
&= e^{\ln a^x} \cdot \ln a \\ \\
&= a^x \ln a \quad \cmark
\end{align*} By the way, notice that this result still applies when $a = e:$
\begin{align*}
\dfrac{d}{dx}e^x &= e^x \cancelto{1}{\ln(e)} \\[8px]
&= e^x
\end{align*} You probably have firmly in mind that $\dfrac{d}{dx}e^x = e^x,$ and so just need to remember that if the base of the exponential a is anything other than e, you have to multiply the derivative by $\ln a.$ For instance, $\dfrac{d}{dx}3^x = 3^x \ln 3.$

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[hide solution]

Let’s next use the Chain Rule as an alternate way to develop the Quotient Rule, starting from the Product Rule.

Practice Problem #2: Quotient Rule from Chain and Product Rules

Use the Chain Rule and the Product Rule to develop the Quotient Rule. Start from $\dfrac{f(x)}{g(x)} = f(x) \Big( g(x)\Big)^{-1}$.

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[hide solution]

And let’s find some more common trig derivatives.

Practice Problem #3: Derivatives of sec and cot

Derive the following, as requested.

(a) Find the derivative of $\sec \theta = \dfrac{1}{\cos \theta}$. Hint: Remember the Chain Rule.

(b) Find the derivative of $\cot \theta = \dfrac{\cos \theta}{\sin \theta} = (\cos \theta)(\sin \theta)^{-1}$. Use the Product Rule.

(c) Find the derivative of $\cot \theta = \dfrac{\cos \theta}{\sin \theta}$. Use the Quotient Rule.

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Since we’re trying to show something about the derivative of an even function, start with its definition and take the derivative:
\begin{align*}
f(-x) &= f(x) \\ \\
\frac{d}{dx}\left[f(-x)\right] &= \frac{d}{dx}\left[f(x)\right] \\ \\
f'(-x) \cdot \left[ \frac{d}{dx} (-x) \right] &= f'(x) \\ \\
f'(-x)\cdot (-1) &= f'(x) \\ \\
– f'(-x) &= f'(x) \\ \\
f'(-x) &= -f'(x)
\end{align*}
The last line shows, as requested, that the derivative of $f$, $f’$, is itself an odd function: $f'(-x) = -f'(x)$. $\quad \cmark$ A less formal approach, using “stuff”: We can view the function $f(-x)$ as being $f(\text{stuff})$, where $\text{“stuff”} = -x$. Then \begin{align*}
f(\text{stuff}) &= f(x) \\ \\
\frac{d}{dx}\left[f(\text{stuff})\right] &= \frac{d}{dx}\left[f(x)\right] \\ \\
f'(\text{stuff}) \cdot \left[ \frac{d}{dx}\text{(stuff)} \right] &= f'(x) \\ \\
f'(-x) \cdot \left[ \frac{d}{dx} (-x) \right] &= f'(x) \\ \\
f'(-x)\cdot (-1)&= f'(x) \\ \\
f'(-x) &= -f'(x)
\end{align*}
The last line shows, as requested, that the derivative of $f$, $f’$, is itself an odd function: $f'(-x) = -f'(x)$. $\quad \cmark$

Since we’re trying to show something about the derivative of an odd function, start with its definition and take the derivative:
\begin{align*}
f(-x) &= -f(x) \\ \\
\frac{d}{dx}\left[f(-x)\right] &= \frac{d}{dx}\left[-f(x)\right] \\ \\
f'(-x) \cdot \left[ \frac{d}{dx} (-x) \right] &= -f'(x) \\ \\
f'(-x)\cdot (-1) &= -f'(x) \\ \\
f'(-x) &= f'(x)
\end{align*}
The last line shows, as requested, that the derivative of $f$, $f’$, is itself an even function: $f'(-x) = f'(x)$. $\quad \cmark$ A less formal approach, using “stuff”: We can view the function $f(-x)$ as being $f(\text{stuff})$, where $\text{“stuff”} = -x$. Then \begin{align*}
f(\text{stuff}) &= -f(x) \\ \\
\frac{d}{dx}\left[f(\text{stuff})\right] &= \frac{d}{dx}\left[-f(x)\right] \\ \\
f'(\text{stuff}) \cdot \left[ \frac{d}{dx}\text{(stuff)} \right] &= -f'(x) \\ \\
f'(-x) \cdot \left[ \frac{d}{dx} (-x) \right] &= -f'(x) \\ \\
f'(-x)\cdot (-1)&= -f'(x) \\ \\
f'(-x) &= f'(x)
\end{align*}
The last line shows, as requested, that the derivative of $f$, $f’$, is itself an even function: $f'(-x) = f'(x)$. $\quad \cmark$

[hide solution]

This screen concludes our focus on the Chain Rule, though of course we’ll be using it often as we proceed.

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