# Derivative Definition (The Derivative as a Function)

Motivating the concept of the derivative is an essential step in a student’s calculus education. This article goes through this definition carefully and with several examples allowing a beginning student to absorb the information. Finding the tangent line equation is then detailed, followed by an important theorem: differentiability implies continuity.

We begin with the definition of the derivative as a limit of a difference quotient. We then give several examples of how to find the derivative of a function using this definition. Finding an equation of the tangent line is then considered, and after several examples of this, we then give examples of how a function may not be differentiable. At the end we prove that every function that is differentiable at a point must also be continuous at that point.

## The Definition of Derivative

The first main idea of calculus is of course, the limit. A limiting process can be used in the study of curves in general; but the derivative is the main limiting process that has lead to the development of calculus.

Given a function $f$ of a real variable $x$ and a positive change in $x$, say $\Delta x$, the expression \begin{equation} \frac{f(x+\Delta x)-f(x)}{\Delta x} \end{equation} is called the difference quotient and is the formula for the slope of a secant line to the graph of $f$ through the points $(x,f(x))$ and $(x+\Delta x,f(x+\Delta x)).$

The limiting process illustrated in the examples below was first developed by the French mathematician Pierre de Fermat. The following definition was realized by Newton and Leibniz.

Definition. The derivative of a function $y=f(x)$ for any $x$ is defined by \begin{equation} f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \end{equation} provided this limit exists.

The derivative is also denoted by $$\frac{d y}{d x}$$ and other common notations are $y’$, $\displaystyle \frac{d f}{d x}$ and $D_x(y).$ Also the limit is sometimes denoted by \begin{equation} \frac{d y}{d x}=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}=f'(x). \end{equation} The process of finding the derivative is called differentiation. A function $f$ is called differentiable at $x$ when the defining limit exists; and we say that $f$ is differentiable on $(a,b)$ when $f$ is differentiable at every point in the open interval $(a,b).$

## Taking the Derivative Using the Definition

In the following examples we illustrate how to find the derivative function using the definition of the derivative.

Example. Find the derivative of the function using the definition of the derivative given $f(x)=x^3$ at $(1,1).$

Solution. By definition we compute the limit, as follows, \begin{align} f'(x) & = \lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} \\ & = \lim_{\Delta x\to 0}\frac{x^3+3 x^2 (\Delta x)+3 x (\Delta x)^2+(\Delta x)^3-x^3}{\Delta x} \\ & = \lim_{\Delta x\to 0}\frac{3 x^2 (\Delta x)+3 x (\Delta x)^2+(\Delta x)^3}{\Delta x} \\ & = \lim_{\Delta x\to 0}\left( 3 x^2 +3 x (\Delta x)+(\Delta x)^2 \right) \\ & = 3 x^2 \end{align} which is the derivative of $f$ for any $x.$ At $x=1,$ we have $f'(1)=1.$

Example. Find the derivative of the function using the definition of the derivative given $f(x)=\sqrt{x}$ at $(8,2).$

Solution. By the definition of the derivative, \begin{align} f'(x) & = \lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} \\ & =\lim_{\Delta x\to 0}\frac{\sqrt{(x+\Delta x)}-\sqrt{x}}{\Delta x} \\ & =\lim_{\Delta x\to 0}\left(\frac{\sqrt{(x+\Delta x)}-\sqrt{x}}{\Delta x}\right)\left(\frac{\sqrt{(x+\Delta x)^2}+\sqrt{(x+\Delta x)x}+\sqrt{x^2}}{\sqrt{(x+\Delta x)^2}+\sqrt{(x+\Delta x)x}+\sqrt{x^2}}\right) \\ & =\lim_{\Delta x\to 0}\left(\frac{x+\Delta x-x}{\Delta x\left(\sqrt{(x+\Delta x)^2}+\sqrt{(x+\Delta x)x}+\sqrt{x^2}\right)}\right) \\ & =\lim_{\Delta x\to 0}\left(\frac{1}{\sqrt{(x+\Delta x)^2}+\sqrt{(x+\Delta x)x}+\sqrt{x^2}}\right) \\ & =\lim_{\Delta x\to 0}\frac{1}{3\sqrt{x^2}} \end{align} which is the derivative of $f$ for any $x.$ At $x=8,$ we have $f'(8)=1/12.$

## Finding an Equation of the Tangent Line

Theorem. The slope of the tangent line to the function at $\left(x_0, f\left(x_0\right) \right)$ is $m=f’\left(x_0\right)$ and $$y=f’\left(x_0\right)\left(x-x_0\right)+f\left(x_0\right)$$ is an equation of the tangent line to the graph of $f$ at $x=x_0.$

Proof. By definition, the slope of the tangent line at $x_0$ is $f'(x_0)$, therefore an equation of the tangent line that passes through $\left(x_0,f\left(x_0\right)\right)$ is $$y=f’\left(x_0\right)\left(x-x_0\right)+f\left(x_0\right)$$ as needed.

Example. Differentiate the function $\displaystyle f(x)=3\sqrt{-x}$ and find the slope of the tangent line at the point where $x=-2.$

Solution. To find the slope of the tangent line we determine the derivative of the function, \begin{align} f'(x) & =\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} \\ & =\lim_{\Delta x\to 0}\frac{3\sqrt{-(x+\Delta x)}-3\sqrt{-x}}{\Delta x} \\ & =\lim_{\Delta x\to 0}\left(3\frac{\sqrt{-(x+\Delta x)}-\sqrt{-x}}{\Delta x}\right)\left(\frac{\sqrt{-(x+\Delta x)}+\sqrt{-x}}{\sqrt{-(x+\Delta x)}+\sqrt{-x}}\right) \\ & =\lim_{\Delta x\to 0}\left(3\frac{-(x+\Delta x)-(-x)}{\Delta x\left(\sqrt{-(x+\Delta x)}+\sqrt{-x}\right)}\right) \\ & =\lim_{\Delta x\to 0}\left(3\frac{-\Delta x}{\Delta x\left(\sqrt{-(x+\Delta x)}+\sqrt{-x}\right)}\right) \\ & =\lim_{\Delta x\to 0}\left(\frac{-3}{\sqrt{-(x+\Delta x)}+\sqrt{-x}}\right) \\ & =-\frac{3}{2 \sqrt{-x}} \end{align} which is the derivative of the function $f$ for any $x<0.$ At $x=-2,$ we have \begin{equation} f'(-2)=-\frac{3}{2 \sqrt{-(-2)}}=-\frac{3\sqrt{2}}{4}. \end{equation} Therefore, an equation of the tangent line at $x=-2$ is
\begin{align} & y=f’\left(x_0\right)\left(x-x_0\right)+f\left(x_0\right) =-\frac{3\sqrt{2}}{4} (x+2)+3\sqrt{2}. \end{align}
as desired.

Example. Find an equation of the tangent line to the graph of \begin{equation}
g(x)=\frac{1-x}{2+x} \end{equation} at $x=-1.$

Solution. To find the slope of the tangent line we compute the derivative of the function,
\begin{align} g'(x) & =\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} \\ & =\lim_{\Delta x\to 0}\frac{\frac{1-(x+\Delta x)}{2+(x+\Delta x)}-\frac{1-x}{2+x}}{\Delta x} \\ & =\lim_{\Delta x\to 0}\frac{\left(\frac{1-(x+\Delta x)}{2+(x+\Delta x)}\right)\left(\frac{2+x}{2+x}\right)-\left(\frac{1-x}{2+x}\right)\left(\frac{2+(x+\Delta x)}{2+(x+\Delta x)}\right)}{\Delta x} \\ & =\lim_{\Delta x\to 0}\frac{-3\Delta x}{\Delta x(2+(x+\Delta x))(2+x)} \\ & =\lim_{\Delta x\to 0}\frac{-3}{(2+(x+\Delta x))(2+x)} \\ & =\frac{-3}{(2+x)^2} \end{align} At $x=-1,$ we have \begin{equation}
g'(-1)=\frac{-3}{(2+(-1))^2}=-3. \end{equation} Therefore, an equation of the tangent line at $x=-1$ is \begin{equation} y=f’\left(x_0\right)\left(x-x_0\right)+f\left(x_0\right) = -3 (x+1)+2. \end{equation} as needed.

## Examples of Non-differentiable Functions

Example. Give three examples of functions $f$ where $f$ is not differentiable at $x=c$ but $f$ is continuous at $x=c.$

Solution. The function $f(x)=|x|$ is not differentiable at $x=0$ since \begin{equation} \lim_{h\to 0^-}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0^-}\frac{|h|}{h}=-1 \end{equation} \begin{equation} \lim_{h\to 0^+}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0^+}\frac{|h|}{h}=1 \end{equation} which proves the two-sided limit (the derivative) \begin{equation} \lim_{h\to 0}\frac{f(0+h)-f(0)}{h} \end{equation} does not exist. This type of example where the function is not differentiable is called a corner point.

Secondly, the function $f(x)=\sqrt{x^2}$ is not differentiable at $x=0$ since \begin{equation} \lim_{h\to 0^-}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0^-}\frac{\sqrt{h^2}}{h} =\frac{1}{\sqrt{h}}=-\infty \end{equation} \begin{equation} \lim_{h\to 0^+}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0^+}\frac{\sqrt{h^2}}{h} =\frac{1}{\sqrt{h}}=+\infty \end{equation} which proves the two-sided limit (the derivative) \begin{equation} \lim_{h\to 0}\frac{f(0+h)-f(0)}{h} \end{equation} does not exist. This type of example where the function is not differentiable is called a vertical tangent.

Thirdly, the function \begin{equation} f(x)= \begin{cases} x-1 & \text{ if } x<0 \\ 2x & \text{ if } x\geq 0 \end{cases} \end{equation} is not differentiable at $x=0$ since \begin{equation} \lim_{h\to 0^-}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0^-}\frac{h}{h}=1 \end{equation} \begin{equation} \lim_{h\to 0^+}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0^+}\frac{2h}{h}=2 \end{equation} which proves the two-sided limit (the derivative) \begin{equation} \lim_{h\to 0}\frac{f(0+h)-f(0)}{h} \end{equation} does not exist. In this third example, does $f$ have a corner point at $x=0$ or is it a vertical tangent at $x=0$?

Example. Compute the difference quotient for the function defined by \begin{equation} f(x)= \begin{cases} \frac{\sin x}{x} & \text{ if } x\neq 0 \\ 1 & \text{ if } x=0 \end{cases} \end{equation} Do you think $f$ is differentiable at $x=0$? If so, what is the equation of the tangent line at $x=0$?

Solution. For $x\neq 0,$ we find, \begin{equation} f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0 }\frac{\frac{\sin (x+h)}{x+h}-\frac{\sin x}{x}}{h}. \end{equation} At $x=0$ we have, \begin{equation} f'(0)=\lim_{h\to 0 }\frac{\frac{\sin (h)}{h}-1}{h}. \end{equation} Using a table of values to compute the limit we infer that $f'(0)=0$ and so the equation of the tangent line is $y=1.$

## Differentiability Implies Continuity

Theorem. If a function $f$ is differentiable at $x=c,$ then it is also continuous at $x=c.$

Proof. Assume that $f$ is a differentiable function, then by definition\begin{equation}
f'(c) =\lim_{\Delta x\to 0}\frac{f(c+\Delta x)-f(c)}{\Delta x} \end{equation} exists, and therefore we can use the product rule for limits with \begin{align} & \lim_{\Delta x\to 0}f(c+\Delta x)-f(c) \\ & \qquad =\lim_{\Delta x\to 0}\frac{f(c+\Delta x)-f(c)}{\Delta x}\Delta x \\ &\qquad =\left( \lim_{\Delta x\to 0}\frac{f(c+\Delta x)-f(c)}{\Delta x} \right)\left( \lim_{\Delta x\to 0}\Delta x \right)\\ &\qquad =f'(c)(0) \\ & \qquad =0 \end{align} Whence, \begin{equation} \lim_{\Delta x\to 0}f(c+\Delta x)=f(c)\end{equation} and so $f$ is continuous at $x=c.$

## Exercises on Derivative Definition

Exercise. Using the definition of the derivative find the derivative for each of the following functions.

$(1) \quad \displaystyle f(x)=2\sqrt{x}-3$

$(2) \quad \displaystyle g(x)=\frac{1}{2\sqrt{x}-3}$

$(3) \quad \displaystyle f(x)=e^{2x+1}$

$(4) \quad \displaystyle h(x)=(x+1)^2$

Exercise. Find an equation of the tangent line at the indicated point for each of the following:

$(1) \quad \displaystyle f(x)=3\sqrt{-x}$ at $x=-2$

$(2) \quad \displaystyle f(z)=\frac{1-z}{2 z}$ at $z=-1$

$(3) \quad \displaystyle f(x)=4-x^2$ at $x=1$

$(4) \quad \displaystyle y=2x^3$ at $x=2$

Exercise. (a) If possible, give an example of a function that is continuous on $(-\infty ,+\infty )$ but does not have a derivative at $x=2.$ Sketch and explain. (b) If possible, give an example of a function that is continuous and increasing on $(-\infty ,2),$ continuous and decreasing on $(2,+\infty),$ discontinuous at $x=2,$ and does not have a derivative at $x=2.$ Sketch and explain.

Exercise. Using the definition find the derivative function given \begin{equation} g(x)= \begin{cases} -2x+3 & \text{if}\text{ }x<\frac{3}{2} \\ -\left(x-\frac{3}{2}\right)^2 & \text{if }x\geq \frac{3}{2} \end{cases} \end{equation}

Exercise. Using the definition find the derivative for each of the following functions.

$(1) \quad \displaystyle v(t)=t-\frac{1}{t}$

$(2) \quad \displaystyle f(x)=(x+1)^3$

$(3) \quad \displaystyle s(t)=1-3t^2$

$(4) \quad \displaystyle w(z)=z+\sqrt{z}$

Exercise. Using the alternative definition for the derivative of a function: \begin{equation} f'(x) =\lim_{z\to x}\frac{f(z)-f(x)}{z-x} \end{equation} find the derivative of each of the following functions.

$(1) \quad \displaystyle f(x)=\frac{x}{x-1}$

$(2) \quad \displaystyle g(x)=\frac{x-1}{x+1}$

$(3) \quad \displaystyle h(x)=(x+1)^3$

$(4) \quad \displaystyle p(t)=2-3t^2$ David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.