The derivative at a point

The instantaneous rate of change at $a$ is the value the average rates of change over $[a, b]$ approach as $b$ moves towards $a$. So far we have estimated that value from a shrinking interval. We can now define it exactly.

The interval from $a$ to $b$ is what is shrinking. Call this $h = b - a$, which means $b = a + h$. Sliding $b$ towards $a$ is then the same as letting $h$ shrink towards $0$. The average rate of change over the interval $[a, b] = [a, a+h]$ then becomes

which is called the difference quotient. It is the secant slope for a step of size $h$, written in terms of that step.

The instantaneous rate of change is the value this difference quotient approaches as $h \to 0$. That value, taken as a limit, is the derivative of $f$ at $a$.

For $f(x) = x^2 - 6x$, simplify the difference quotient at $a = 4$ to a single expression in $h$.

Solution

The difference quotient at $a = 4$ is

We expand $f(4 + h)$:

Since $f(4) = 16 - 24 = -8$, the numerator is

Every term in the numerator carries a factor of $h$, so we cancel it against the $h$ in the denominator:

The difference quotient for $f(x) = x^2 - 6x$ at $a = 4$ is $2 + h$.

The derivative $f'(a)$ is the limit of the difference quotient $\dfrac{f(a+h) - f(a)}{h}$ as $h \to 0$. Computing it means evaluating that limit.

We cannot evaluate the limit by setting $h = 0$ in the difference quotient directly: that gives $\dfrac{0}{0}$, which has no value. Simplifying the difference quotient first clears this, leaving an expression where letting $h \to 0$ is an ordinary substitution.

The result is a single number: the exact instantaneous rate of change of $f$ at $a$, and the slope of the tangent line there.

For $f(x) = x^2 + 3x$, compute $f'(2)$ from the limit definition.

Solution

We are working at $a = 2$, so the derivative is

We form the difference quotient, expanding $f(2+h)$. Replacing $x$ with $2+h$ in $f(x) = x^2 + 3x$:

The fixed value is $f(2) = 2^2 + 3(2) = 10$. Substituting both into the difference quotient gives

Now we simplify the numerator so that every remaining term carries a factor of $h$. The constants have already cancelled, and $h^2 + 7h = h(h + 7)$:

Substituting $h = 0$ at this stage would give $\dfrac{0}{0}$, so we cancel the common factor $h$ first, which is valid because $h \neq 0$ as it approaches $0$:

Finally we let $h \to 0$ in the simplified expression:

So $f'(2) = 7$: the instantaneous rate of change of $f$ at $x = 2$ is $7$.

The tangent line to the graph of $f$ at $(a, f(a))$ has slope $f'(a)$, the derivative at $a$. Knowing the slope and a point on the line, we can write down its equation.

On that line, the slope between the point of tangency $(a, f(a))$ and any other point $(x, y)$ must equal $f'(a)$. This can be written as:

and multiplying through by $x - a$ gives its equation.

The two ingredients are the slope $f'(a)$ and the point $(a, f(a))$. Both come from the function and the value $a$.

The tangent line equation can be left in this form, or rearranged into $y = mx + c$.

For $f(x) = x^2 - 4x$, find the equation of the tangent line at $x = 1$.

Solution

The tangent line needs two ingredients: its slope $f'(1)$ and the point of tangency $(1, f(1))$.

The point of tangency is $(1, -3)$, since $f(1) = 1 - 4 = -3$. The slope is the derivative $f'(1) = -2$.

Now we substitute the slope $f'(1) = -2$ and the point $(1, -3)$ into the tangent line equation:

which gives:

We rearrange into slope-intercept form:

The tangent line to $f$ at $x = 1$ is $y = -2x - 1$.