From a secant to a tangent

What you will be able to do

• explain the derivative as the limit of the slopes of secants;
• find a derivative from the definition, for a power and for a square root;
• differentiate any polynomial and write the equation of a tangent.

How fast is it going right now?

A car covers $100$ km in two hours. Its average speed is $50$ km/h. The speedometer needs more: it shows the speed at one instant. An instant has no length, so we cannot divide a distance by a time of zero. The way out is the limit. Take a very short time, work out the average speed over it, and see what number the averages approach.

The same idea gives the slope of a curve. A slope is easy for a straight line. For a curve, the slope changes from point to point, so we ask for the slope at a point.

The secant and the tangent

Let $P = (a, f(a))$ be a point of the graph. Let $Q = (a + h, f(a + h))$ be another point, a little to the right. The line through $P$ and $Q$ is a secant. Its slope is

Now let $Q$ slide along the curve towards $P$. Then $h$ gets smaller and the secant turns. In the limit it becomes the tangent, the line that just touches the curve at $P$.

The definition

The derivative of $f$ at $a$ is the limit of the secant slopes:

If the limit exists, we say that $f$ is differentiable at $a$. Putting $x = a + h$ gives the same limit in another form, $f'(a) = \displaystyle\lim_{x \to a} \dfrac{f(x) - f(a)}{x - a}$. This form is the standard limit $\dfrac{x^n - a^n}{x - a}$ for $f(x) = x^n$.

The derivative has several names. The slope of the tangent is called the gradient in the UK and the slope in the US. The symbols $f'(x)$, $\dfrac{dy}{dx}$ and $\dfrac{d}{dx}f(x)$ mean the same thing. The meaning is a rate: how many units of $y$ change for one unit of $x$.

Your first derivative, from the definition

Take $f(x) = x^2$ and find $f'(x)$.

1. Write the quotient: $\dfrac{(x + h)^2 - x^2}{h}$.
2. Expand the top: $(x + h)^2 - x^2 = 2xh + h^2$.
3. Divide by $h$: the quotient is $2x + h$.
4. Let $h \to 0$: the limit is $2x$.

So $f'(x) = 2x$. At $x = 3$ the tangent has slope $6$.

The basic rules

The same method gives these results. Each one is a limit that you can check.

• A constant. If $f(x) = c$, then $f(a + h) - f(a) = 0$, so $f'(x) = 0$.
• A power. If $f(x) = x^n$, then $f'(x) = nx^{n-1}$. This is the power limit $\dfrac{x^n - a^n}{x - a} \to na^{n-1}$ from the last course, and it works for fractions and negative $n$ too.
• A constant multiple. $(cf)' = c\,f'$.
• A sum. $(f + g)' = f' + g'$.

For example, $\dfrac{d}{dx}\left(4x^3 - 5x^2 + 7x - 2\right) = 12x^2 - 10x + 7$.

A square root by the definition. For $f(x) = \sqrt{x}$ the quotient is $\dfrac{\sqrt{x + h} - \sqrt{x}}{h}$. Multiply the top and the bottom by $\sqrt{x + h} + \sqrt{x}$. The top becomes $h$, so the quotient is $\dfrac{1}{\sqrt{x + h} + \sqrt{x}}$, and the limit is $\dfrac{1}{2\sqrt{x}}$. This agrees with the power rule for $n = \dfrac{1}{2}$.

The tangent line

At the point $(a, f(a))$ the tangent has slope $f'(a)$. Its equation is $y - f(a) = f'(a)(x - a)$.

For $y = x^3$ at $a = 2$: $f(2) = 8$ and $f'(2) = 3 \times 4 = 12$. The tangent is $y - 8 = 12(x - 2)$, which is $y = 12x - 16$.

Where the Class 12 chapter goes further

Not every function has a derivative at every point. The graph of $|x|$ has a corner at $0$: the slopes of the secants are $-1$ on the left and $1$ on the right, so the limit does not exist. The Class 12 chapter on continuity and differentiability studies this. It also adds the chain rule, and implicit and logarithmic differentiation, which let you differentiate $\sin 2x$ and $x^x$. This course stays with the definition and the basic rules.

Practise with new numbers

Try it in Python

The program prints the slope of the secant for $f(x) = x^2$ between $x = 3$ and $x = 3 + h$ for smaller and smaller $h$. Then it does the same for $x^3$ at $x = 2$.

Common mistakes

• Dividing by $h = 0$. The quotient is defined for $h \ne 0$. Simplify first, then let $h \to 0$.
• Forgetting that the secant slope needs both points. The slope is $f(a + h) - f(a)$ over $h$, not $f(a + h)$ over $h$.
• Writing $f'(x)$ when you mean $f'(a)$. The first is a function. The second is one number.
• Lowering the power but forgetting the factor. $\dfrac{d}{dx}x^5 = 5x^4$, not $x^4$.

Remember these

• $f'(a) = \displaystyle\lim_{h \to 0} \dfrac{f(a + h) - f(a)}{h}$ is the slope of the tangent at $x = a$.
• $\dfrac{d}{dx}(c) = 0$, $\dfrac{d}{dx}x^n = nx^{n-1}$, $(cf)' = cf'$, $(f + g)' = f' + g'$.
• The tangent at $a$ is $y - f(a) = f'(a)(x - a)$.
• A corner means no derivative.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Use the definition to find the derivative of $f(x) = 3x^2$.

2. For $f(x) = x^2$, find the slope of the secant through $x = 2$ and $x = 2.1$. Compare it with the slope of the tangent at $x = 2$.

3. Differentiate $5x^4 - 3x^2 + 2x - 7$.

4. Find the equation of the tangent to $y = x^3$ at $x = 2$.

5. Find the derivative of $\sqrt{x}$ at $x = 9$ from the definition or from the power rule.

6. Is $f(x) = |x|$ differentiable at $x = 0$?

References

• OpenStax, Calculus Volume 1, Chapter 3, sections 3.1 and 3.2 on defining the derivative and the derivative as a function.
• NCERT, Mathematics Textbook for Class XI, Chapter 12, the sections on derivatives and the algebra of derivatives.