1.2: The Derivative- Limit Approach

The Derivative: Limit Approach

The following definition generalizes the example from the previous section (concerning instantaneous velocity) to a general function \(f(x)\):

For a general function \(f(x)\), the derivative \(f'(x)\) represents the instantaneous rate of change of \(f\) at \(x\), i.e. the rate at which \(f\) changes at the “instant” \(x\). For the limit part of the definition only the intuitive idea of how to take a limit—as in the previous section—is needed for now. Notice that the above definition makes the derivative \(f'\) itself a function of the variable \(x\). The function \(f'\) can be evaluated at specific values of \(x\), or you can write its general formula \(f'(x)\).

The (instantaneous) velocity of an object as the derivative of the object’s position as a function of time is only one physical application of derivatives. There are many other examples:

The limit definition can be used for finding the derivatives of simple functions.

Example \(\PageIndex{1}\): derivconst

Add text here.

Solution

Find the derivative of the function \(f(x) = 1\).

Solution: By definition, \(f(x) = 1\) for all \(x\), so:

\[\begin{aligned} f'(x) ~&=~ \lim_{\Delta x \to 0} ~\frac{f(x+\Delta x) ~-~ f(x)} {\Delta x}\

\[6pt] &=~ \lim_{\Delta x \to 0} ~\frac{1 ~-~ 1}{\Delta x}\

\[6pt] &=~ \lim_{\Delta x \to 0} ~\frac{0}{\Delta x}\

\[6pt] &=~ \lim_{\Delta x \to 0} ~0\

\[4pt] f'(x) ~&=~ 0 \end{aligned} \nonumber \]

Notice in the above example that replacing \(\Delta x\) by \(0\) was unnecessary when taking the limit, since the ratio \(\frac{f(x + \Delta x) ~-~ f(x)}{\Delta x}\) simplified to 0 before taking the limit, and the limit of 0 is 0 regardless of what \(\Delta x\) approaches. In fact, the answer—namely, \(f'(x) = 0\) for all \(x\)—should have been obvious without any calculations: the function \(f(x) = 1\) is a constant function, so its value (1) never changes , and thus its rate of change is always 0. Hence, its derivative is 0 everywhere. Replacing the constant 1 by any constant yields the following important result:

The above discussion shows that the calculation in Example

Example \(\PageIndex{1}\): derivconst

Add text here.

Solution

was unnecessary. Consider another example where no calculation is required to find the derivative: the function \(f(x) = x\). The graph of this function is just the line \(y = x\) in the \(xy\)-plane, and the rate of change of a line is a constant, called its slope. The line \(y = x\) has a slope of 1, so the derivative of \(f(x) = x\) is \(f'(x) = 1\) for all \(x\). The formal calculation of the derivative, though unnecessary, verifies this:

\[f'(x) ~=~ \lim_{\Delta x \to 0} ~\frac{f(x+\Delta x) ~-~ f(x)}{\Delta x} ~=~ \lim_{\Delta x \to 0} ~\frac{(x + \Delta x) ~-~ x}{\Delta x} ~=~ \lim_{\Delta x \to 0} ~\frac{\Delta x}{\Delta x} ~=~ \lim_{\Delta x \to 0} ~1 ~=~ 1 \nonumber \]

Recall that a function whose graph is a line is called a linear function. For a general linear function \(f(x) = mx + b\), where \(m\) is the slope of the line and \(b\) is its \(y\)-intercept, the same argument as above for \(f(x) = x\) yields the following result:

The function \(f(x) = 1\) from Example

Example \(\PageIndex{1}\): derivconst

Add text here.

Solution

is the special case where \(m = 0\) and \(b = 1\); its graph is a horizontal line, so its slope (and hence its derivative) is 0 for all \(x\). Likewise, the function \(f(x) = 2x - 1\) represents a line of slope \(m = 2\), so its derivative is 2 for all \(x\). Figure [fig:derivlines] shows these and other linear functions \(y=f(x)\).

Linear functions have a constant derivative—the constant being the slope of the line. The converse turns out to be true: a function with a constant derivative must be a linear function.¹¹ What types of functions do not have constant derivatives? The previous section discussed such a function: the parabola \(s(t) = -16t^2 + 100\), whose derivative \(s'(t) = -32t\) is clearly not a constant function. In general, functions that represent curves (i.e. not straight lines) do not change at a constant rate—that is precisely what makes them curved. So such functions do not have a constant derivative.

Find the derivative of the function \(f(x) = \dfrac{1}{x}\). Also, find the instantaneous rate of change of \(f\) at \(x=2\).

Solution: For all \(x \ne 0\), the derivative \(f'(x)\) is:

\[\begin{aligned} f'(x) ~&=~ \lim_{\Delta x \to 0} ~\frac{f(x+\Delta x) ~-~ f(x)} {\Delta x}\

\[6pt] &=~ \lim_{\Delta x \to 0} ~\frac{~\dfrac{1}{x + \Delta x} ~-~ \dfrac{1}{x}~} {\Delta x} ~\rightarrow~ \frac{0}{0}\quad\text{, so simplify the ratio before plugging in $\Delta x = 0$,}\

\[6pt] &=~ \lim_{\Delta x \to 0} ~\frac{~\dfrac{x ~-~ (x + \Delta x)} {(x + \Delta x)x}~}{\Delta x} \quad\text{(after getting a common denominator)}\

\[6pt] &=~ \lim_{\Delta x \to 0} ~\frac{-\cancel{\Delta x}}{\cancel{\Delta x}(x + \Delta x)x}\

\[6pt] &=~ \lim_{\Delta x \to 0} ~\frac{-1}{(x + \Delta x)x} ~=~ \frac{-1}{(x+0)x}\

\[6pt] f'(x) ~&=~ -\frac{1}{x^2} \end{aligned} \nonumber \]

The instantaneous rate of change of \(f\) at \(x=2\) is just the derivative \(f'(x)\) evaluated at \(x=2\), that is, \(f'(2) = -\frac{1}{2^2} = -\frac{1}{4}\).

Notice that the instantaneous rate of change \(f'(2) = -\frac{1}{4}\) in the above example is a negative number. This should make sense, since the function \(f(x) = \frac{1}{x}\) is changing in the negative direction at \(x=2\); that is, \(f(x)\) is decreasing in value at \(x=2\). This is plain to see from the graph of \(f(x) = \frac{1}{x}\) shown on the right. In fact, for all \(x \ne 0\) the function \(f(x) = \frac{1}{x}\) is decreasing as \(x\) grows.¹² This is reflected in the derivative \(f'(x) = -\frac{1}{x^2}\) being negative for all \(x \ne 0\). In general, a negative derivative means that the function is decreasing, while a positive derivative means that it is increasing. The problem with using the limit definition to find the derivative of a curved function is that the calculations require more work, as the above example shows. As the functions become more complicated those calculations can become difficult or even impossible. And though limits have not yet been defined formally, for now the intuitively obvious idea of limits suffices, namely:

Below are some simple rules for limits, which will be proved later:

The above rules say that the limit of sums, differences, constant multiples, products, and quotients is the sum, difference, constant multiple, product, and quotient, respectively, of the limits. This seems intuitively obvious.

These rules can be used for finding other expressions for the derivative. The quantity \(\Delta x\) represents a small number—positive or negative—that approaches 0, but it is common in mathematics texts to use the letter \(h\) instead:¹³

\[\label{eqn:hderivative} \setlength{\fboxsep}{4pt}\boxed{f'(x) ~=~ \lim_{h \to 0} ~\frac{f(x + h) ~-~ f(x)}{h}} \]

Another formulation is to set \(h=w-x\) in formula ([eqn:hderivative]), which yields

\[f'(x) ~=~ \lim_{h \to 0} ~\frac{f(x + h) ~-~ f(x)}{h} ~=~ \lim_{w-x \to 0} ~\frac{f(x + (w-x)) ~-~ f(x)}{w ~-~ x} ~, \nonumber \]

so that

\[\label{eqn:wxderivative} \setlength{\fboxsep}{4pt}\boxed{f'(x) ~=~ \lim_{w \to x} ~\frac{f(w) ~-~ f(x)}{w ~-~ x}} \]

since \(w-x\) approaches 0 if and only if \(w\) approaches \(x\). Another formulation replaces \(h\) by \(-h\):

\[f'(x) ~=~ \lim_{h \to 0} ~\frac{f(x + h) ~-~ f(x)}{h} ~=~ \lim_{-h \to 0} ~\frac{f(x + -h) ~-~ f(x)}{-h} ~=~ \lim_{-h \to 0} ~\frac{-\left(f(x) ~-~ f(x - h)\right)}{-h}~, \nonumber \]

and thus

\[\label{eqn:neghderivative} \setlength{\fboxsep}{4pt}\boxed{f'(x) ~=~ \lim_{h \to 0} ~\frac{f(x) ~-~ f(x-h)}{h}} \]

since \(-h\) approaches 0 if and only if \(h\) approaches \(0\). The above formulations did not use the Limit Rules, but the following result does:

Since \(f'(x) ~=~ \displaystyle\lim_{h \to 0} ~\dfrac{f(x + h) ~-~ f(x)} {h} ~=~ \displaystyle\lim_{h \to 0} ~\dfrac{f(x) ~-~ f(x-h)}{h}\) by formulas ([eqn:hderivative]) and ([eqn:neghderivative]), then Limit Rule (c) shows that

\[\frac{1}{2}f'(x) ~=~ \lim_{h \to 0} ~\frac{f(x + h) ~-~ f(x)} {2h} ~=~ \lim_{h \to 0} ~\frac{f(x) ~-~ f(x-h)}{2h} ~. \nonumber \]

Now use the idea that \(a - b = (a - c) + (c - b)\) for all \(a\), \(b\), and \(c\) to write:

\[\begin{aligned} \lim_{h \to 0} ~\frac{f(x+h) ~-~ f(x-h)}{2h} ~&=~ \lim_{h \to 0} ~\frac{\left(f(x+h) ~-~ f(x)\right) ~+~ \left(f(x) ~-~ f(x-h)\right)}{2h}\

\[6pt] &=~ \lim_{h \to 0} ~\frac{f(x+h) ~-~ f(x)}{2h} ~+~ \lim_{h \to 0} ~\frac{f(x) ~-~ f(x-h)}{2h} \quad\text{(by Limit Rule (a))}\

\[6pt] &=~ \frac{1}{2} \cdot f'(x) ~+~ \frac{1}{2} \cdot f'(x)\

\[6pt] &=~ f'(x)\end{aligned} \nonumber \]

As an example of using these different formulations, recall that a function \(f\) is even if \(f(-x) = f(x)\) for all \(x\) in the domain of \(f\), and \(f\) is odd if \(f(-x) = -f(x)\) for all \(x\) in its domain. For example, \(x^2\), \(x^4\), and \(\cos\,x\) are even functions; \(x\), \(x^3\), and \(\sin\,x\) are odd functions. The following result is often useful:

To prove the first statement—the second is an exercise—suppose that \(f\) is an even function and that \(f'(x)\) exists for all \(x\) in its domain. Then

\[\begin{aligned} {3} f'(-x) ~&=~ \lim_{h \to 0} ~\frac{f(-x + h) ~-~ f(-x)}{h} \qquad&&\text{by formula (\ref{eqn:hderivative}) with $x$ replaced by $-x$}\

\[6pt] &=~ \lim_{h \to 0} ~\frac{f(-(x - h)) ~-~ f(-x)}{h} &&{}\

\[6pt] &=~ \lim_{h \to 0} ~\frac{f(x - h) ~-~ f(x)}{h} &&\text{since $f$ is even}\

\[6pt] &=~ \lim_{h \to 0} ~\frac{-\left(f(x) ~-~ f(x-h)\right)}{h} &&{}\

\[6pt] &=~ -\lim_{h \to 0} ~\frac{f(x) ~-~ f(x-h)}{h} &&\text{by Limit Rule (c), so}\

\[6pt] f'(-x) ~&=~ -f'(x) &&\text{by formula (\ref{eqn:neghderivative}),}\end{aligned} \nonumber \]

which shows that \(f'\) is an odd function.

Derivatives do not always exist, as the following example shows.

Example \(\PageIndex{1}\): absnondiff

Add text here.

Solution

Let \(f(x) = \abs{x}\). Show that \(f'(0)\) does not exist.

Solution: Recall that the absolute value function \(f(x) = \abs{x}\) is defined as

\[f(x) ~=~ \abs{x} ~=~ \begin{cases} \phantom{~-}x & \text{if } x \ge 0\\ ~-x & \text{if } x < 0 \end{cases} \nonumber \]

The graph consists of two lines meeting at the origin. For \(x \ge 0\) the graph is the line \(y = x\), which has slope 1. For \(x \le 0\) the graph is the line \(y = -x\), which has slope -1. These lines agree in value (\(y=0\)) at \(x=0\), but their slopes do not agree in value at \(x=0\). Therefore the derivative of \(f\) does not exist at \(x=0\), since the derivative of a curve is just its slope. A more “formal” proof (which amounts to the same argument) is outlined in the exercises.

If the derivative \(f'(x)\) exists then \(f\) is differentiable at \(x\). A differentiable function is one that is differentiable at every point in its domain. For example, \(f(x) = x\) is a differentiable function, but \(f(x) = \abs{x}\) is not differentiable at \(x=0\). The act of calculating a derivative is called differentiation. For example, differentiating the function \(f(x) = x\) yields \(f'(x) = 1\).

[sec1dot2]

Note: For all exercises, you can use anything discussed so far (including previous exercises).

For Exercises 1-11, find the derivative of the given function \(f(x)\) for all \(x\) (unless indicated otherwise).

4

\(f(x) = 0\)

\(f(x) = 1 - 3x\)

\(f(x) = (x+1)^2\)

\(f(x) = 2x^2 - 3x + 1\)

3

\(f(x) = \frac{1}{x+1}\), for all \(x \ne -1\)

\(f(x) = \frac{-1}{x+1}\), for all \(x \ne -1\)

\(f(x) = \frac{1}{x^2}\), for all \(x \ne 0\)

[exer:sqrtderiv] \(f(x) = \sqrt{x}\), for all \(x > 0~\) (Hint: Rationalize the numerator in the definition of the derivative.)

3

\(f(x) = \sqrt{x+1}\), for all \(x > -1\)

\(f(x) = \sqrt{x^2 + 1}\)

\(f(x) = \sqrt{x^2 + 3x + 4}\)

In Exercise [exer:sqrtderiv] the point \(x=0\) was excluded when calculating \(f'(x)\), even though \(x=0\) is in the domain of \(f(x) = \sqrt{x}\). Can you explain why \(x=0\) was excluded? [[1.]]

Show that for all functions \(f\) such that \(f'(x)\) exists, \(f'(x) ~=~ \displaystyle\lim_{w \to x} ~\dfrac{f(x) ~-~ f(w)}{x ~-~ w} ~\).

True or false: If \(f\) and \(g\) are differentiable functions on an interval \((a,b)\) and \(f(x) < g(x)\) for all \(x\) in \((a,b)\), then \(f'(x) < g'(x)\) for all \(x\) in \((a,b)\). If true, prove it; if false, give a counterexample. Would your answer change if the restriction of \(x\) to \((a,b)\) were removed and all real \(x\) were used instead?

Show that the derivative of an odd function is an even function.

For Exercises [exer:altderivfirst]-[exer:altderivlast], assuming that \(f'(x)\) exists, prove the given formula. [[1.]]

2

[exer:altderivfirst] \(f'(x) ~=~ \displaystyle\lim_{h \to 0} ~\dfrac{f(x + 2h) ~-~ f(x - 2h)}{4h} ~\)

\(f'(x) ~=~ \displaystyle\lim_{h \to 0} ~\dfrac{f(x + 3h) ~-~ f(x - 3h)}{6h} ~\)

2

\(f'(x) ~=~ \displaystyle\lim_{h \to 0} ~\dfrac{f(x + 2h) ~-~ f(x - 3h)}{5h} ~\)

\(f'(x) ~=~ \displaystyle\lim_{h \to 0} ~\dfrac{f(x + ah) ~-~ f(x - bh)}{(a+b)h} ~\quad\)(\(a,b>0\))

2

\(\displaystyle\lim_{w \to x} ~\dfrac{w\,f(x) ~-~ x\,f(w)}{w ~-~ x} ~=~ f(x) ~-~ x\,f'(x)\)

[exer:altderivlast] \(\displaystyle\lim_{w \to x} ~\dfrac{w^2\,f(x) ~-~ x^2\,f(w)}{w ~-~ x} ~=~ 2x\,f(x) ~-~ x^2\,f'(x)\)

Show that \(f(x) = \abs{x}\) is not differentiable at \(x=0\), using formula ([eqn:hderivative]) for the derivative. Here you will have to use a part of the definition which has not been used yet: as \(h\) approaches 0, \(h\) can be either positive or negative. Consider those two cases in showing that the limit is not defined at \(x=0\).

Suppose that \(f(a+b) = f(a)f(b)\) for all \(a\) and \(b\), and \(f'(0)\) exists. Show that \(f'(x)\) exists for all \(x\).