We now define the “derivative” explicitly, based on the limiting slope ideas of the previous section. Then we see how to compute some simple derivatives.

Let us now generalise what we did in the last section so as to find “the slope of the curve \(y=f(x)\) at \((x_0,y_0)\)” for any smooth enough ^{ 1 } function \(f(x)\text{.}\)

As before, let \((x_0,y_0)\) be any point on the curve \(y=f(x)\text{.}\) So we must have \(y_0=f(x_0)\text{.}\) Now let \((x_1,y_1)\) be any other point on the same curve. So \(y_1=f(x_1)\) and \(x_1\ne x_0\text{.}\) Think of \((x_1,y_1)\) as being pretty close to \((x_0,y_0)\) so that the difference

\begin{gather*}
\De x=x_1-x_0
\end{gather*}

in \(x\)–coordinates is pretty small. In terms of this \(\De x\) we have

\begin{gather*}
x_1=x_0+\De x\qquad\text{and}\qquad
y_1=f\big(x_0+\De x\big)
\end{gather*}

We can construct a secant line through \((x_0,y_0)\) and \((x_1,y_1)\) just as we did for the parabola above. It has slope

\begin{gather*}
\frac{y_1-y_0}{x_1-x_0}=\frac{f\big(x_0+\De x\big)-f(x_0)}{\De x}
\end{gather*}

If \(f(x)\) is reasonably smooth ^{ 2 }, then as \(x_1\) approaches \(x_0\text{,}\) i.e. as \(\De x\) approaches \(0\text{,}\) we would expect the secant through \((x_0,y_0)\) and \((x_1,y_1)\) to approach the tangent line to the curve \(y=f(x)\) at \((x_0,y_0)\text{,}\) just as happened in Figure 2.1.6. And more importantly, the slope of the secant through \((x_0,y_0)\) and \((x_1,y_1)\) should approach the slope of the tangent line to the curve \(y=f(x)\) at \((x_0,y_0)\text{.}\)

Thus we would expect ^{ 3 } the slope of the tangent line to the curve \(y=f(x)\) at \((x_0,y_0)\) to be

\begin{gather*}
\lim_{\De x\rightarrow 0}\frac{f\big(x_0+\De x\big)-f(x_0)}{\De x}
\end{gather*}

When we talk of the “slope of the curve” at a point, what we really mean is the slope of the tangent line to the curve at that point. So “the slope of the curve \(y=f(x)\) at \((x_0,y_0)\)” is also the limit ^{ 4 } expressed in the above equation. The derivative of \(f(x)\) at \(x=x_0\) is also defined to be this limit. Which leads ^{ 5 } us to the most important definition in this text:

######
Definition 2.2.1 Derivative at a point

Let \(a\in\bbbr\) and let \(f(x)\) be defined on an open interval ^{ 6 } that contains \(a\text{.}\)

Lets now compute the derivatives of some very simple functions. This is our first step towards building up a toolbox for computing derivatives of complicated functions — this process will very much parallel what we did in Chapter 1 with limits. The two simplest functions we know are \(f(x)=c\) and \(g(x)=x\text{.}\)

######
Example 2.2.2 Derivative of \(f(x)=c\)

Let \(a, c \in \bbbr\) be a constants. Compute the derivative of the constant function \(f(x) = c\) at \(x=a\text{.}\)

We compute the desired derivative by just substituting the function of interest into the formal definition of the derivative.

\begin{align*}
f'(a) &= \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} && \text{(the definition)}\\
&= \lim_{h \to 0} \frac{c - c}{h} && \text{(substituted in the function)}\\
&= \lim_{h \to 0} 0 &&\text{(simplified things)}\\
&= 0
\end{align*}

That was easy! What about the next most complicated function — arguably it's this one:

######
Example 2.2.3 Derivative of \(g(x)=x\)

Let \(a\in \bbbr\) and compute the derivative of \(g(x) = x\) at \(x=a\text{.}\)

Again, we compute the derivative of \(g\) by just substituting the function of interest into the formal definition of the derivative and then evaluating the resulting limit.

\begin{align*}
g'(a) &= \lim_{h \to 0} \frac{g(a+h) - g(a)}{h}
&& \text{(the definition)}\\
&= \lim_{h \to 0} \frac{(a+h) - a}{h}
&& \text{(substituted in the function)}\\
&= \lim_{h \to 0} \frac{h}{h} && \text{(simplified things)}\\
&= \lim_{h \to 0} 1 && \text{(simplified a bit more)}\\
&= 1
\end{align*}

That was a little harder than the first example, but still quite straight forward — start with the definition and apply what we know about limits.

Thanks to these two examples, we have our first theorem about derivatives:

######
Theorem 2.2.4 Easiest derivatives

Let \(a,c \in \mathbb{R}\) and let \(f(x) = c\) be the constant function and \(g(x) = x\text{.}\) Then

\begin{align*}
f'(a) &= 0\\
\end{align*}

and

\begin{align*}
g'(a) &= 1.
\end{align*}

To ratchet up the difficulty a little bit more, let us redo the example we have already done a few times \(f(x)=x^2\text{.}\) To make it a little more interesting let's change the names of the function and the variable so that it is not exactly the same as Examples 2.1.2 and 2.1.5.

######
Example 2.2.5 Derivative of \(h(t)=t^2\)

Compute the derivative of

\begin{align*}
h(t) &= t^2 & \text{ at } t = a
\end{align*}

- This function isn't quite like the ones we saw earlier — it's a function of \(t\) rather than \(x\text{.}\) Recall that a function is a rule which assigns to each input value an output value. So far, we have usually called the input value \(x\text{.}\) But this “\(x\)” is just a dummy variable representing a generic input value. There is nothing wrong with calling a generic input value \(t\) instead. Indeed, from time to time you will see functions that are not written as formulas involving \(x\text{,}\) but instead are written as formulas in \(t\) (for example representing time — see Section 1.2), or \(z\) (for example representing height), or other symbols.
- So let us write the definition of the derivative
\begin{align*}
f'(a) &= \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}
\end{align*}

and then translate it to the function names and variables at hand:
\begin{align*}
h'(a) &= \lim_{h \to 0} \frac{h(a+h)-h(a)}{h}
\end{align*}

- But there is a problem — “\(h\)” plays two roles here — it is both the function name and the small quantity that is going to zero in our limit. It is extremely dangerous to have a symbol represent two different things in a single computation. We need to change one of them. So let's rename the small quantity that is going to zero in our limit from “\(h\)” to “\(\De t\)”:
\begin{align*}
h'(a) &= \lim_{\De t \to 0} \frac{h(a+\De t)-h(a)}{\De t}
\end{align*}

- Now we are ready to begin. Substituting in what the function \(h\) is,
\begin{align*}
h'(a) &= \lim_{\De t \to 0} \frac{(a+\De t)^2-a^2}{\De t}\\
&= \lim_{\De t \to 0} \frac{a^2+2a\,\De t+\De t^2-a^2}{\De t}
&& \big(\text{just squared out $(a+\De t)^2$}\big)\\
&= \lim_{\De t \to 0} \frac{2a\,\De t+\De t^2}{\De t}\\
&= \lim_{\De t \to 0} (2a +\De t)\\
&= 2a
\end{align*}

- You should go back check that this is what we got in Example 2.1.5 — just some names have been changed.