AP Topics: 2.7, 3.1, 3.5
✎ p169 1–6, 7–33 odd, 41, 45, 49, 53, 57, 61, 65, 69, 73, 77, 81, 85, 89, 93, 97, 99–102, 104, 105, 108, 110, 112, 115, 117, 119, 120, 127, 130, 131, 133 | 141, 143, 159–164, 167, 193–195
Finding the derivative $\frac{dy}{dx}$ means finding out the rate of change of $y$ with respect to $x$, or to put it another way, how $y$ changes as $x$ changes.
Well, what if there was intermediate between $y$ and $x$? We see this with composite functions. Let’s look at a simple one: $y=3(2x)$. Here, we have an inner function $u=2x$, meaning our outer function can be rewritten as $y=3u$. So, in order to see how $y$ changes with respect to $x$, we first need to see how $y$ changes with respect to $u$.
OK, do $dy/du=3$. Whatever is going on with $u$, the rate of change of $y$ will be 3. And $u$ itself changes, with $du/dx=2$. So, no matter where we are on this function, $y$ changes by 3 for each increase in $u$, which in turn increases by 2 for each increase in $x$, meaning $dy/dx=6$.
Well, that should make sense because our composite function was really just $6x$, so the slope is going to be 6, no matter where we look. This is a very simple example with constant rates of change, but the concept applies to the more complicated derivatives.
For another example, for $y=(2x)^2$, let $u=2x$ and $y=u^2$. Now, $dy/du=2u$, so the rate of change will depend on the value of $u$, but it will be double it no matter what. Like earlier, $du/dx=2$, a constant rate of change of 2.
This time, we’ll have to start with $x$ and work our way towards $y$. Let’s say $x=3$. $du/dx$ is still 2, so that doesn’t change. $dy/du=2u=2(2x)$, which in this case will be 12. So, when $x=3$, then $dy/dx=24$. For every 1 in $x$, $u$ is double that, and for every 1 in $u$, $y$ is twelve times that.
\[\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}\]This is the Chain Rule. To find the derivative of a composite function, find the derivative of your outer function with respect to the inner one, and multiply by the derivative of the inner function with respect to $x$.
And here is the alternate way of writing the definition of the Chain Rule.
\[\frac{d}{dx}[f\left(g\left(x\right)\right)] = f'\left(g\left(x\right)\right)g'(x)\]Like with earlier sections, the book has a slew of examples that you should look at. We’ll cover some in class. The book also has rewritten version of your derivative rules that take into account the Chain Rule.
The proof of this, although it uses the Chain Rule, likely won’t make much sense until we cover the next section. So, for now
\[\frac{d}{dx}[\ln x] = \frac{1}{x}\]or accounting for the Chain Rule
\[\frac{d}{dx}[\ln u] = \frac{u'}{u}\]Something to keep in mind is that $\ln$ is undefined for negative numbers, so you might run into $\ln|u|$ with the absolute value there as a safeguard. You can pretend the absolute value is not there when differentiating as the results are the same.
\[\frac{d}{dx}[\ln |u|] = \frac{u'}{u}\]$e^x$ is the inverse function of $\ln x$. So that means $e^{\ln x}=x$. Let’s take that and apply it to an exponential function in another base, reason being is that we know the derivative of $e^x$ already.
\[4^x = e^{(\ln 4)x}\]or more generally
\[a^x = e^{(\ln a)x}\]With the Chain Rule, and keeping in mind that $\ln a$ is just a constant, we can find the derivative of $a^x$.
\[\frac{d}{dx}[a^x] = \frac{d}{dx}[e^{(\ln a)x}]=e^{(\ln a)x} \cdot \ln a = a^x \cdot \ln a\] \[\frac{d}{dx}[a^x] = a^x \cdot \ln a\]And we can do something similar to logarithms with other bases. The base change rule is
\[\log_a b = \frac{\log_c b}{\log_c a}\]where $c$ is whatever new base we want. Well, we want the base to be $e$ since we know that rule already.
\[\log_a x = \frac{\ln x}{\ln a} = \frac{1}{\ln a} \ln x\]Remember that $1/\ln a$ is just a constant as we derive this one.
\[\frac{d}{dx}[\log_a x] = \frac{d}{dx}\left[\frac{1}{\ln a} \ln x\right] = \frac{1}{(\ln a) x}\]So, our two new rules, written with Chain rule in mind are
\[\begin{align*} \frac{d}{dx}[a^u] &= (a^u\ln a)u' \\ \frac{d}{dx}[\log_a u] &= \frac{u'}{u\ln a} \end{align*}\]