5.2 Definition of Derivative

Consider finding a line tangent to a function $f(x)$ that is not a straight line. Specifically, suppose that $f(x)=x^2$, and we want to find an equation of the tangent line to the function at $a$. How can we obtain such a function? Suppose we want to change the average rate of change (which is the slope of a line) between some arbitrary value $x$ and $a$ such that $x\ne a$. Then we have:

$m=\frac{f(x)-f(a)}{x-a}$

The graph of this function is shown below, where the orange line is the average rate of change $m$:

We can find the tangent line to $f(x)$ at the point ($a$, $f(a)$) as we find the limit of $m$ as $x$ approaches $a$. Formally, the tangent line to a continuous function $f(x)$ at the point ($a$, $f(a)$) is the line through ($a$, $f(a)$) with the slope

$m=\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}$

For example, let’s say we have $f(x)=x^2$, and the tangent line goes through the point (2, 4). Then, the slope of the tangent line is

$\begin{array}{rl}m& =\underset{x\to 2}{lim}\frac{f(x)-f(2)}{x-2}\\ & =\underset{x\to 2}{lim}\frac{x^2-2^2}{x-2}\\ & =\underset{x\to 2}{lim}\frac{(x+2)(x-2)}{x-2}\\ & =\underset{x\to 2}{lim}(x+2)\\ & =4\end{array}$

Now the equation of tangent line, given that the slope is $m=4$ and the line passes through (2, 4), is

$y-4=4(x-2)$

We can rewrite this in slope-intercept form as follows:

$y=4x-4$

There is another way to express the slope of a tangent line. Let $h=x-a$. We can rewrite the equation of the slope of the tangent line as follows:

$m=\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}$

This is because as $x$ approaches $a$, $h$ approaches 0. Also, h=x-a means that $x=h+a$.

Suppose $f(x)=\frac{1}{x}$ , and we are looking for the tangent line that goes through (1, 1). Then,

$\begin{array}{rl}m& =\underset{h\to 0}{lim}\frac{f(1+h)-f(1)}{h}\\ & =\underset{h\to 0}{lim}\frac{\frac{1}{1+h}-1}{h}\\ & =\underset{h\to 0}{lim}\frac{\frac{1-(1+h)}{1+h}}{h}\\ & =\underset{h\to 0}{lim}\frac{-h}{h(1+h)}\\ & =\underset{h\to 0}{lim}\frac{-1}{1+h}\\ & =-1\end{array}$

Given that the slope is $m=-1$, and the line passes through (1, 1), the equation of the tangent line is

$y-1=-1(x-1)$

Therefore, in slope-intercept form, we have

$y=-x+2$

The graph of $f(x)$ and its tangent line looks like this:

You may be wondering, “So what is a derivative?” The derivative is a way to show the rate of change of a function $f(x)$. It is the slope of a line that lies tangent to $f(x)$ at a specific point. Formally, the derivative of a function $f(x)$ at a constant $a$ that is denoted by $f^{\prime }(a)$ is

$f^{\prime }(a)=\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}=\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}$

A function $f(x)$ is differentiable at $a$ if $f^{\prime }(a)$ exists. If $f(x)$ is a function, then the function $g(x)=f^{\prime }(x)$ is called the derivative of $f(x)$. The derivative of a function $f(x)$ is sometimes denoted as $\frac{df(x)}{dx}$  next to $f^{\prime }(a)$. We can also write $\frac{d}{dx}$  followed by some formula involving $x$ instead of explicitly naming a function $f(x)$. Mathematically, the definition of the derivative is or we calculate the derivative of a function with the limit by

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

Let's find the derivative of $f(x)=\frac{1}{x}$ .

$\begin{array}{rl}f\prime (x)& =\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}\\ & =\underset{h\to 0}{lim}\frac{\frac{x-(x+h)}{\mathrm{x}(x+h)}}{h}\\ & =\underset{h\to 0}{lim}\frac{-h}{hx(x+h)}\\ & =\underset{h\to 0}{lim}\frac{-1}{x(x+h)}\\ & =-\frac{1}{x^2}\end{array}$

Since we have formally defined a derivative, here is the formal definition of a tangent line. The tangent line to $f(x)$ at a point ($a$, $f(a)$) is the line through ($a$, $f(a)$) whose slope is $f^{\prime }(a)$.