The Derivative as a Function

A derivative at one point is a number. A derivative function assigns such numbers to all inputs where the required finite limit exists. For $f(x)=x^2$, the calculation $f'(a)=2a$ holds for every real $a$, so the derivative function is $f'(x)=2x$. The variable has changed its role from a fixed base point in one proof to the argument of a new function.

The notation $dy/dx$ also denotes the derivative when $y=f(x)$. At this stage it is notation for a limit-defined rate, not permission to cancel arbitrary symbols. Later, the chain rule and substitution will explain why certain manipulations that resemble fraction arithmetic work.

Cubes and integer powers

For $f(x)=x^3$, the difference quotient is

$$\frac{(x+h)^3-x^3}h=3x^2+3xh+h^2,\qquad h\ne0.$$

Letting $h\to0$ gives $f'(x)=3x^2$. Both positive and negative nonzero increments yield the same expression’s limit.

For a positive integer $n$, the binomial theorem gives

$$(x+h)^n=x^n+nx^{n-1}h+\sum_{k=2}^n {n\choose k}x^{n-k}h^k.$$

Subtract $x^n$ and divide by nonzero $h$. Every term of the sum contains a positive power of $h$ and tends to zero. The result is the power rule $d(x^n)/dx=nx^{n-1}$ for positive integers. The case $n=1$ is simply a constant quotient $1$. A constant function has derivative zero because its output difference is always zero. Extending the power rule to all real exponents needs additional results; the binomial argument alone does not supply that extension.

Reciprocal and square root

The reciprocal calculation gives

$$\left(\frac1x\right)'=-\frac1{x^2},\qquad x\ne0.$$

The derivative cannot include zero because the original function has no value there. For $f(x)=\sqrt{x}$ at $x>0$, conjugates give

$$\frac{\sqrt{x+h}-\sqrt{x}}h =\frac1{\sqrt{x+h}+\sqrt{x}} \longrightarrow\frac1{2\sqrt{x}}.$$

The function’s domain is $[0,\infty)$, while its ordinary two-sided derivative has domain $(0,\infty)$. Even the right-hand difference quotient at zero grows without bound, so it does not supply a finite endpoint derivative. A formula and its domain must travel together.

For $|x|$, the derivative is $-1$ on $(-\infty,0)$ and $1$ on $(0,\infty)$, and is undefined at zero. The derivative function records a change of slope and the missing derivative at the corner. Drawing a line between the two branches would add information the function does not contain.

Higher derivatives and interpretation

If $f'$ is differentiable, its derivative is $f''$, the second derivative. For $f(x)=x^3$, $f'=3x^2$, $f''=6x$, $f'''=6$, and every derivative of order four or greater is zero. The $n$th derivative is written $f^{(n)}$, while $f^n$ normally means a power; these symbols are not interchangeable.

For position $s(t)$, the first derivative is velocity and the second acceleration. Positive acceleration means velocity is increasing, but speed need not increase. If a vehicle is moving in the negative direction with $v=-10$ and $a=2$, its velocity moves toward zero and its speed initially decreases. The signs of velocity and acceleration together determine this behavior.

Reading a derivative graph

Where $f'>0$, tangent slopes are positive; where $f'<0$, they are negative. A theorem in a later chapter will justify conclusions about increase on an interval. A derivative equal to zero at one point does not itself identify a maximum or minimum: $x^3$ has derivative zero at zero and continues increasing through it.

Two different functions can have the same derivative. For instance $x^2$ and $x^2+7$ both have derivative $2x$. Differentiation discards a vertical offset. The mean value theorem will show that, on an interval, functions with the same derivative can differ only by a constant. This observation will become the foundation of antiderivatives.

Exercises

1. Differentiate $x^5$ directly by expansion, identifying the term that survives the limit.
2. Give the domains of $\sqrt{x-2}$ and its derivative.
3. Find the first three derivatives of $x^4$.
4. Describe the derivative of $|x-3|$ at every real input where it exists.
5. Give two functions with derivative $4x^3$.
6. Explain why negative velocity with negative acceleration produces increasing speed locally when both signs persist.

Worked solutions

1. The quotient is $5x^4+10x^3h+10x^2h^2+5xh^3+h^4$. Only $5x^4$ survives.
2. The function is defined on $[2,\infty)$; the finite ordinary derivative $1/(2\sqrt{x-2})$ is defined on $(2,\infty)$.
3. $4x^3$, $12x^2$, and $24x$.
4. It equals $-1$ for $x<3$, $1$ for $x>3$, and is undefined at $3$ because the one-sided slopes disagree.
5. $x^4$ and $x^4+10$ are examples.
6. Velocity becomes more negative; its magnitude therefore increases. Speed is $-v$ while $v<0$, so its derivative is $-a>0$.