More Derivatives¶

The graph $y=x$ is a straight line with slope 1.

We can also plug in $f(x)=x$ to the definition of derivative:

We can also use the $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ thing with $y=x$:

The graph $y=c$ is a horizontal line at height $c$, so its slope is $0$.

We can also plug in $f(x)=c$ to the definition of derivative:

We can use the $\displaystyle\lim_{\Delta x \to 0} \tfrac{\Delta y}{\Delta x}$ thing by setting $y=c$. Then $\Delta y = 0$ because $y$ is a constant; its value doesn't change.

Let's plug in $c \cdot f(x)$ to the definition of derivative.

Here's a nice consequence of this rule.

We did this in the previous chapter:

If these things work, then the stuff above is correct:

• $\frac{d}{dx}(f(x)+g(x))=f'(x)+g'(x)$
• $\frac{d}{dx}(f(x)-g(x))=f'(x)-g'(x)$

This is because adding and substracting a big pile of stuff can be always turned into simpler adds and substractions...

...and then we could apply the above rules to one thing at a time, starting with $a+b$.

Let's start by proving the $+$ thing with the definition of derivative. It's kind of messy, but it works. I'll color $f$'s with blue and $g$'s with green so the mess should be a bit less unreadable.

We can prove the minus thing using stuff that we have proved earlier:

Let's first assume that $g(x+h) \ne g(x)$ with a small $h$; that is, $g(x+h)-g(x) \ne 0$ in a $h \to 0$ limit. This means that we can divide stuff by $\green{g(x+h)-g(x)}$. We'll do the $g(x+h)=g(x)$ case separately.

Multiplying top and bottom by the green stuff is a nice and clever way to get started with this:

The stuff with green on the bottom looks kind of like something that could be turned into the definition of derivative, but the defintion has just $h$ on the bottom instead of the green mess. Let's define another variable $\Delta g = \green{g(x+h)-g(x)}$. We assumed that the derivative of $g$ exists, so $g$ must be continuous. So if $h \to 0$, then $g(x+h) \to g(x)$ and $\bigl(\green{g(x+h)-g(x)}\bigr) \to 0$.

If $g(x+h)=g(x)$ when $h \to 0$, then we can just replace $g(x+h)$ with $g(x)$ everywhere:

We are now done with the proof, but I also want to show you what this rule looks like with the $\displaystyle\lim_{\Delta x \to 0}$ thing. For convenience, we'll abbreviate $f(g(x))$ as $f$ and $g(x)$ as $g$. Now $f'(x) = \displaystyle\lim_{\Delta x\to0}\frac{\Delta f}{\Delta x}$, but note that $f'(\maroon{g(x)}) = \displaystyle\lim_{\Delta g \to 0}\frac{\Delta f}{\Delta\maroon g}$ because we're putting $g(x)$ into $f'(\quad)$.

It looks like the $\Delta g$'s just cancel out, and in fact, the green stuff we added in the beginning of the proof was $\Delta g$. This is how I knew that adding it would do something useful.

Also note that this notation doesn't handle the corner case with $g$ being a constant because then $\green{\Delta g} = 0$ and $\frac{\Delta f}{\Delta g}$ is dividing by zero.

Plugging into the definition of derivative gives this:

If the top was slightly different, we could just group it nicely.

We have $-f(x)g(x)$ instead of $-f(x)g(x+h)$, but we can just put a $-f(x)g(x+h)$ there if we also add $+f(x)g(x+h)$ to cancel it out. The blue parts add up to $0$:

In the previous chapter we checked the $x^2$ derivative with a fun square area thing, and we can do the same thing here if we assume that $f(x)$ and $g(x)$ are positive. The area of an $f(x)$ by $g(x)$ rectangle is $f(x)g(x)$.

Now taking $\Delta x \to 0$ limits gives $\dfrac{d(f(x)g(x))}{dx} = f'(x)g(x) + f(x)g'(x) + f'(x)\cdot\red0$ where $\red0=\displaystyle\lim_{\Delta x \to 0}\red{\Delta g}$.

It's easy to prove that this works for an individual $c$ value. For example, our very first derivative example shows that $\frac{d}{dx} x^2 = 2x$, and plugging in $c=2$ gives $2x^{2-1}=2x^1=2x$. Let's prove that this works with all positive integers instead of just 2 using a powerful technique known as induction.

Let's start by showing that this works with $c=1$:

Next we'll prove that if the rule works at $c=k$ then it also works at $c=k+1$ where $k$ is a positive integer. Let's write things down just to be clear:

We assume: $\frac{d}{dx} x^k = k x^{k-1}$

We'll prove: $\frac{d}{dx} x^{k+1} = (k+1)x^{(k+1)-1}$

Let's use the $\frac{d}{dx}(f(x)g(x))$ and $\frac{d}{dx} x$ rules we proved above and the assumption.

We proved that if $\frac{d}{dx} x^k = k x^{k-1}$ then $\frac{d}{dx} x^{k+1} = (k+1)x^{(k+1)-1}$. Now we know that the rule works when $c=1$, and then if we plug in $k=1$ we know it works when $c=2$, and so on.

Note that we only proved that the rule works when $c$ is a positive integer, but it also works when $c$ is e.g. $\frac{1}{2}$. This tutorial also contains another proof that doesn't have this restriction.

We could prove this with the $\frac{d}{dx} x^c$ rule because $\sqrt x = x^{1/2}$, but we proved the $x^c$ rule only for positive integers. Let's survive without it.

You're probably wondering how the heck I knew that multiplying top and bottom by the green stuff would result in something useful, but this is actually a well-known trick with square roots and limits.

We could use the $\frac{d}{dx} x^c$ rule again because $1/x$ is actually $x^{-1}$, but as before, we haven't proved that it works at $c=-1$ so we'll do this without it. Adding the green and blue stuff means that the bottoms are the same and we can combine stuff together.

The $-1$ is kind of surprising, but as you can see, it comes from the $-\frac 1 x$ on the top. $x^2$ is never negative so $\frac{1}{x^2}$ is also never negative, and $\frac{-1}{x^2}$ is always negative (it can't be $0$ because $x$ is finite). So it seems like $\frac 1 x$ is decreasing; that is, its values always get smaller as $x$ gets bigger.

>>> 1/2
0.5
>>> 1/3
0.3333333333333333
>>> 1/4
0.25
>>> 1/5
0.2

Let's see what the definition of derivative gives for $\frac{d}{dx}|x|$ at $x=0$:

But $h\to0$ means that we must get the same answer from both cases. The limit does not exist.

Let's check with the definition of derivative.

If $h\to0^+$, $1/h$ is dividing by something very small and positive, so it goes to $\infty$. Similarly, in the second case $h$ is negative and $1/h$ goes to $-\infty$... but $-1/h$ goes to $\infty$:

Most mathematicians say that the derivative cannot be $\infty$ or $-\infty$ because many ways to use derivatives only work with finite values. In fact, limits at infinity are a kind of corner case anyway, and some people mean "the limit exists and is finite" when they say "the limit exists".

Let's update our definition of derivative: