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The derivative function

We now know that the instantaneous rate of change of a function f(x) at x = a, or equivalently the slope of the tangent line to the graph of y = f(x) at x = a, is given by the value f'(a).

Introduction

We now know that the instantaneous rate of change of a function \(f(x)\) at \(x = a\), or equivalently the slope of the tangent line to the graph of \(y = f(x)\) at \(x = a\), is given by the value \(f'(a)\). In all of our examples so far, we have identified a particular value of \(a\) as our point of interest: \(a = 1\), \(a = 3\), etc. But it is not hard to imagine that we will often be interested in the derivative value for more than just one \(a\)-value, and possibly for many of them. In this section, we explore how we can move from computing the derivative at a single point to computing a formula for \(f'(a)\) at any point \(a\). Indeed, the process of taking the derivative generates a new function, denoted by \(f'(x)\), derived from the original function \(f(x)\).

Exploration

How the derivative is itself a function

In your work in Preview Activity with \(f(x) = 4x - x^2\), you may have found several patterns. One comes from observing that \(f'(0) = 4\), \(f'(1) = 2\), \(f'(2) = 0\), and \(f'(3) = -2\). That sequence of values leads us naturally to conjecture that \(f'(4) = -4\) and \(f'(5) = -6\). We also observe that the particular value of \(a\) has very little effect on the process of computing the value of the derivative through the limit definition. To see this more clearly, we compute \(f'(a)\), where \(a\) represents a number to be named later. Following the now standard process of using the limit definition of the derivative, \[\begin{aligned}f'(a) =\mathstrut \amp \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} = \lim_{h \to 0} \frac{4(a + h) - (a + h)^2 - (4a-a^2)}{h} \\ =\mathstrut \amp \lim_{h \to 0} \frac{4a + 4h - a^2 - 2ha - h^2 - 4a+a^2}{h} = \lim_{h \to 0} \frac{4h - 2ha - h^2}{h} \\ =\mathstrut \amp \lim_{h \to 0} \frac{h(4 - 2a - h)}{h} = \lim_{h \to 0} (4 - 2a - h)\end{aligned}\].

Here we observe that neither \(4\) nor \(2a\) depend on the value of \(h\), so as \(h \to 0\), \((4 - 2a - h) \to (4 - 2a)\). Thus, \(f'(a) = 4 - 2a\).

This result is consistent with the specific values we found above: e.g., \(f'(3) = 4 - 2(3) = -2\). And indeed, our work confirms that the value of \(a\) has almost no bearing on the process of computing the derivative. We note further that the letter being used is immaterial: whether we call it \(a\), \(x\), or anything else, the derivative at a given value is simply given by 4 minus 2 times the value. We choose to use \(x\) for consistency with the original function given by \(y = f(x)\), as well as for the purpose of graphing the derivative function. For the function \(f(x) = 4x - x^2\), it follows that \(f'(x) = 4 - 2x\).

Because the value of the derivative function is linked to the graph of the original function, it makes sense to look at both of these functions plotted on the same domain.

An excellent way to explore how the graph of \(f(x)\) generates the graph of \(f'(x)\) is through an interactive graphic. See, for instance, gvsu.edu/s/5C or gvsu.edu/s/5D, via the sites of David Austin and Marc Renault.

In Section when we first defined the derivative, we wrote the definition in terms of a value \(a\) to find \(f'(a)\). As we have seen above, the letter \(a\) is merely a placeholder, and it often makes more sense to use \(x\) instead. For the record, here we restate the definition of the derivative.

Condensed — the full section is in Boelkins, Active Calculus.

Summary

The limit definition of the derivative, \(f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\), produces a value for each \(x\) at which the derivative is defined, and this leads to a new function \(y = f'(x)\). It is especially important to note that taking the derivative is a process that starts with a given function (\(f\)) and produces a new, related function (\(f'\)).
There is essentially no difference between writing \(f'(a)\) (as we did regularly in Section) and writing \(f'(x)\). In either case, the variable is just a placeholder that is used to define the rule for the derivative function.
Given the graph of a function \(y = f(x)\), we can sketch an approximate graph of its derivative \(y = f'(x)\) by observing that heights on the derivative's graph correspond to slopes on the original function's graph.
In Activity, we encountered some functions that had sharp corners on their graphs, such as the shifted absolute value function. At such points, the derivative fails to exist, and we say that \(f\) is not differentiable there. For now, it suffices to understand this as a consequence of the jump that must occur in the derivative function at a sharp corner on the graph of the original function.

Practice (10)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

In part (a) of , why do you get some positive and some negative values for \(f'(a)\)?
In calculus, there's lots of notation. It's important to think about what various letters and symbols mean. The following questions ask you to think about meaning in the context of derivative notation.
1. In \(f'(a)\), what does \(a\) represent? Is \(a\) an input? an output? a slope? a height? something else?
2. What does \(f'(a)\) represent? Is \(f'(a)\) an input? an output? a slope? a height? something else?
3. What do \(f'(a)\) and \(a\) have to do with each other?
4. What do \(f'(a)\) and \(f(a)\) have to do with each other?
Let \(f\) be a function with the following properties: \(f\) is differentiable at every value of \(x\) (that is, \(f\) has a derivative at every point), \(f(-2) = 1\), and \(f'(-2) = -2\), \(f'(-1) = -1\), \(f'(0) = 0\), \(f'(1) = 1\), and \(f'(2) = 2\).
1. On the axes provided at left in Figure, sketch a possible graph of \(y = f(x)\). Explain why your graph meets the stated criteria.
2. Conjecture a formula for the function \(y = f(x)\). Use the limit definition of the derivative to determine the corresponding formula for \(y = f'(x)\). Discuss both graphical and algebraic evidence for whether or not your conjecture is correct.
Vis svaret
1. The fact that \(f\) is differentiable everywhere means that the graph of \(f\) is smooth everywhere. The slopes of the tangent lines to \(f\) are negative but increasing on the interval \((-\infty,0)\) and positive and increasing on the interval \((0,\infty)\), with a slope of \(0\) when \(x=0\). This is the kind of behavior that a quadratic function possesses, so we could guess that \(f\) has a graph something like that shown in the figure below.
2. Because the change in the derivative values is constant, it looks like \(f'\) is linear with a slope of \(1\), passing through the point \((0,0)\), so it is reasonable to guess that \(f'(x) = x\). A plot of \(f'\) is shown at right in the figure below.
3. A natural guess is \(f(x) = x^2\); since we need the function to pass through the point \((1,-2)\), we might try \(f(x) = x^2 - 3\). Using the limit definition, we have \[\begin{aligned}f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &= \lim_{h \to 0} \frac{\left[(x+h)^2-3\right] - \left[x^2-3\right]}{h} \\ &= \lim_{h \to 0} \frac{\left[x^2+2xh+h^2-3\right] - \left[x^2-3\right]}{h} \\ &= \lim_{h \to 0} \frac{2xh+h^2}{h} \\ &= \lim_{h \to 0} 2x+h \\ &= 2x\end{aligned}\]. So this guess is close, but is off by a factor of \(2\), since we want \(f'(x) = x\). Instead, if we use \(f(x) = \frac{1}{2}x^2-1\) (note that we chose the\(-1\) so that \(f(-2)=1\), then we have \[\begin{aligned}f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &= \lim_{h \to 0} \frac{\left[\frac{1}{2}(x+h)^2-1\right] - \left[\frac{1}{2}x^2-1\right]}{h} \\ &= \lim_{h \to 0} \frac{\frac{1}{2}\left[x^2+2xh+h^2\right] - \frac{1}{2}\left[x^2\right]}{h} \\ &= \lim_{h \to 0} \frac{1}{2}\frac{2xh+h^2}{h} \\ &= \lim_{h \to 0} x+\frac{h}{2} \\ &= x\end{aligned}\]. This appears to be the correct function \(f\).
Consider the function \(g(x) = x^2 - x + 3\).
1. Use the limit definition of the derivative to determine a formula for \(g'(x)\).
2. Use a graphing utility to plot both \(y = g(x)\) and your result for \(y = g'(x)\); does your formula for \(g'(x)\) generate the graph you expected?
3. Use the limit definition of the derivative to find a formula for \(p'(x)\) where \(p(x) = 5x^2 - 4x + 12\).
4. Compare and contrast the formulas for \(g'(x)\) and \(p'(x)\) you have found. How do the constants 5, 4, 12, and 3 affect the results?
Vis svaret
1. By definition, \[\begin{aligned}g'(x) &= \lim_{h \to 0} \frac{g(x+h)-g(x)}{h} \\ &= \lim_{h \to 0} \frac{(x+h)^2 - (x+h) + 3 - (x^2 - x + 3)}{h} \\ &= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x-h + 3 - x^2 + x - 3}{h} \\ &= \lim_{h \to 0} \frac{2xh + h^2 - h}{h} \\ &= \lim_{h \to 0} \frac{h(2x + h - 1)}{h} \\ &= \lim_{h \to 0} (2x + h - 1) \\ &= 2x - 1\end{aligned}\].
2. In the above figure, we see plots of both \(g\) and \(g'\). We observe that the point \((0.5, 2.75)\) is the vertex of the quadratic function \(g\), and at this point the slope of the tangent line to \(g(x)\) is zero. This aligns with the point \((0.5, 0)\) where \(y=g'(x)\) crosses the \(x\)-axis. In addition, we note that \(g'(x)\) is negative for \(x \lt 0.5\), which corresponds to where \(g\) is decreasing and has tangent lines with slopes that are negative. Similarly, the values of \(g'\) are positive for \(x \gt 0.5\), which align with the values of slopes we see on the original function \(g\).
3. By definition, \[\begin{aligned}p'(x) &= \lim_{h \to 0} \frac{p(x+h)-p(x)}{h} \\ &= \lim_{h \to 0} \frac{5(x+h)^2 - 4(x+h) + 12 - (5x^2 - 4x + 12)}{h} \\ &= \lim_{h \to 0} \frac{5x^2 + 10xh + 5h^2 - 4x-4h + 12 - 5x^2 + 4x - 12}{h} \\ &= \lim_{h \to 0} \frac{10xh + 5h^2 - 4h}{h} \\ &= \lim_{h \to 0} \frac{h(10x + 5h - 4)}{h} \\ &= \lim_{h \to 0} (10x + 5h - 4) \\ &= 10x - 4\end{aligned}\].
4. For \(g(x) = x^2 - x + 3\), we found that \(g'(x) = 2x - 1\). For \(p(x) = 5x^2 - 4x + 12\), we determined that \(p'(x) = 10x - 4\). The constants \(3\) and \(12\) don't seem to affect the results at all, and that makes sense because those numbers only serve to shift the graphs of \(g\) and \(p\) vertically, which does nothing to change the slope. The coefficient \(-4\) on the linear term in \(p(x)\) appears to make the \(-4\) appear in \(p'(x)= 10x - 4\). That, too, makes sense in light of the fact that if we considered only the linear function \(L(x) = -4x\), the slope would everywhere be \(-4\), in contrast to the coefficent \(-1\) found on the linear term in \(g(x) = x^2 - x + 3\), which leads to the constant \(-1\) in \(g'(x) = 2x - 1\). Finally, the leading coefficient \(5\) in \((x) = 5x^2 - 4x + 12\) leads to the coefficient of \(10\) in \(p'(x) = 10x -4\). This makes sense because if we considered only \(y = 5x^2\), the \(5\) would make the graph \(5\) times as steep as the graph of \(y = x^2\), and thus it affects the derivative proportionately.
For each graph that provides an original function \(y = f(x)\) in Figure, your task is to sketch an approximate graph of its derivative function, \(y = f'(x)\), on the axes immediately below. View the scale of the grid for the graph of \(f\) as being \(1 \times 1\), and assume the horizontal scale of the grid for the graph of \(f'\) is identical to that for \(f\). If you need to adjust the vertical scale on the axes for the graph of \(f'\), you should label that accordingly.

Vis svaret

At any point where there is a jump in the graph of the derivative, the derivative is undefined. Normally we would draw an open circle at each end of the graph, but those are omitted here for convenience of plotting.
Let \(g\) be a continuous function (that is, one with no jumps or holes in the graph) and suppose that a graph of \(y= g'(x)\) is given by the graph on the right in Figure.
1. Observe that for every value of \(x\) that satisfies \(0 \lt x \lt 2\), the value of \(g'(x)\) is constant. What does this tell you about the behavior of the graph of \(y = g(x)\) on this interval?
2. On what intervals other than \(0 \lt x \lt 2\) do you expect \(y = g(x)\) to be a linear function? Why?
3. At which values of \(x\) is \(g'(x)\) not defined? What behavior does this lead you to expect to see in the graph of \(y=g(x)\)?
4. Suppose that \(g(0) = 1\). On the axes provided at left in Figure, sketch an accurate graph of \(y = g(x)\).
Vis svaret
1. Since \(g'(x)\) is constant (with value \(1\)) on the interval \(0 \lt x \lt 2\), it follows that \(g\) is linear on that same interval, since \(g\) is increasing at a constant rate.
2. On \(-3.5 \lt x \lt -2\), we also expect \(g\) to be linear with slope \(1\), while on \(-2 \lt x \lt 0\) and \(2 \lt x \lt 3.5\), \(g\) will be linear with slope \(-1\); in each case this is true because the value of \(g'(x)\) is constant with the noted value on the interval.
3. From the given graph of \(g'(x)\), we observe that \(g'\) is undefined at \(x = -2, 0, 2\). Since we have been given that \(g\) is a continuous function, we can conclude that \(g\) must have sharp corners on its graph at these points. Moreover, that makes sense in light of our earlier observations that show \(g\) has constant slope on the intervals that connect at \(x = -2, 0, 2\) and the graph jumps from having slope \(1\) to \(-1\), and then back, and so on.
Let \(f\) be a function with the following properties: \(f\) is differentiable at every value of \(x\) (that is, \(f\) has a derivative at every point), \(f(-2) = 1\), and \(f'(-2) = -2\), \(f'(-1) = -1\), \(f'(0) = 0\), \(f'(1) = 1\), and \(f'(2) = 2\).
1. On the axes provided at left in Figure, sketch a possible graph of \(y = f(x)\). Explain why your graph meets the stated criteria.
2. Conjecture a formula for the function \(y = f(x)\). Use the limit definition of the derivative to determine the corresponding formula for \(y = f'(x)\). Discuss both graphical and algebraic evidence for whether or not your conjecture is correct.
Vis svaret
1. The fact that \(f\) is differentiable everywhere means that the graph of \(f\) is smooth everywhere. The slopes of the tangent lines to \(f\) are negative but increasing on the interval \((-\infty,0)\) and positive and increasing on the interval \((0,\infty)\), with a slope of \(0\) when \(x=0\). This is the kind of behavior that a quadratic function possesses, so we could guess that \(f\) has a graph something like that shown in the figure below.
2. Because the change in the derivative values is constant, it looks like \(f'\) is linear with a slope of \(1\), passing through the point \((0,0)\), so it is reasonable to guess that \(f'(x) = x\). A plot of \(f'\) is shown at right in the figure below.
3. A natural guess is \(f(x) = x^2\); since we need the function to pass through the point \((1,-2)\), we might try \(f(x) = x^2 - 3\). Using the limit definition, we have \[\begin{aligned}f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &= \lim_{h \to 0} \frac{\left[(x+h)^2-3\right] - \left[x^2-3\right]}{h} \\ &= \lim_{h \to 0} \frac{\left[x^2+2xh+h^2-3\right] - \left[x^2-3\right]}{h} \\ &= \lim_{h \to 0} \frac{2xh+h^2}{h} \\ &= \lim_{h \to 0} 2x+h \\ &= 2x\end{aligned}\]. So this guess is close, but is off by a factor of \(2\), since we want \(f'(x) = x\). Instead, if we use \(f(x) = \frac{1}{2}x^2-1\) (note that we chose the\(-1\) so that \(f(-2)=1\), then we have \[\begin{aligned}f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &= \lim_{h \to 0} \frac{\left[\frac{1}{2}(x+h)^2-1\right] - \left[\frac{1}{2}x^2-1\right]}{h} \\ &= \lim_{h \to 0} \frac{\frac{1}{2}\left[x^2+2xh+h^2\right] - \frac{1}{2}\left[x^2\right]}{h} \\ &= \lim_{h \to 0} \frac{1}{2}\frac{2xh+h^2}{h} \\ &= \lim_{h \to 0} x+\frac{h}{2} \\ &= x\end{aligned}\]. This appears to be the correct function \(f\).
Consider the function \(g(x) = x^2 - x + 3\).
1. Use the limit definition of the derivative to determine a formula for \(g'(x)\).
2. Use a graphing utility to plot both \(y = g(x)\) and your result for \(y = g'(x)\); does your formula for \(g'(x)\) generate the graph you expected?
3. Use the limit definition of the derivative to find a formula for \(p'(x)\) where \(p(x) = 5x^2 - 4x + 12\).
4. Compare and contrast the formulas for \(g'(x)\) and \(p'(x)\) you have found. How do the constants 5, 4, 12, and 3 affect the results?
Vis svaret
1. By definition, \[\begin{aligned}g'(x) &= \lim_{h \to 0} \frac{g(x+h)-g(x)}{h} \\ &= \lim_{h \to 0} \frac{(x+h)^2 - (x+h) + 3 - (x^2 - x + 3)}{h} \\ &= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x-h + 3 - x^2 + x - 3}{h} \\ &= \lim_{h \to 0} \frac{2xh + h^2 - h}{h} \\ &= \lim_{h \to 0} \frac{h(2x + h - 1)}{h} \\ &= \lim_{h \to 0} (2x + h - 1) \\ &= 2x - 1\end{aligned}\].
2. In the above figure, we see plots of both \(g\) and \(g'\). We observe that the point \((0.5, 2.75)\) is the vertex of the quadratic function \(g\), and at this point the slope of the tangent line to \(g(x)\) is zero. This aligns with the point \((0.5, 0)\) where \(y=g'(x)\) crosses the \(x\)-axis. In addition, we note that \(g'(x)\) is negative for \(x \lt 0.5\), which corresponds to where \(g\) is decreasing and has tangent lines with slopes that are negative. Similarly, the values of \(g'\) are positive for \(x \gt 0.5\), which align with the values of slopes we see on the original function \(g\).
3. By definition, \[\begin{aligned}p'(x) &= \lim_{h \to 0} \frac{p(x+h)-p(x)}{h} \\ &= \lim_{h \to 0} \frac{5(x+h)^2 - 4(x+h) + 12 - (5x^2 - 4x + 12)}{h} \\ &= \lim_{h \to 0} \frac{5x^2 + 10xh + 5h^2 - 4x-4h + 12 - 5x^2 + 4x - 12}{h} \\ &= \lim_{h \to 0} \frac{10xh + 5h^2 - 4h}{h} \\ &= \lim_{h \to 0} \frac{h(10x + 5h - 4)}{h} \\ &= \lim_{h \to 0} (10x + 5h - 4) \\ &= 10x - 4\end{aligned}\].
4. For \(g(x) = x^2 - x + 3\), we found that \(g'(x) = 2x - 1\). For \(p(x) = 5x^2 - 4x + 12\), we determined that \(p'(x) = 10x - 4\). The constants \(3\) and \(12\) don't seem to affect the results at all, and that makes sense because those numbers only serve to shift the graphs of \(g\) and \(p\) vertically, which does nothing to change the slope. The coefficient \(-4\) on the linear term in \(p(x)\) appears to make the \(-4\) appear in \(p'(x)= 10x - 4\). That, too, makes sense in light of the fact that if we considered only the linear function \(L(x) = -4x\), the slope would everywhere be \(-4\), in contrast to the coefficent \(-1\) found on the linear term in \(g(x) = x^2 - x + 3\), which leads to the constant \(-1\) in \(g'(x) = 2x - 1\). Finally, the leading coefficient \(5\) in \((x) = 5x^2 - 4x + 12\) leads to the coefficient of \(10\) in \(p'(x) = 10x -4\). This makes sense because if we considered only \(y = 5x^2\), the \(5\) would make the graph \(5\) times as steep as the graph of \(y = x^2\), and thus it affects the derivative proportionately.
For each graph that provides an original function \(y = f(x)\) in Figure, your task is to sketch an approximate graph of its derivative function, \(y = f'(x)\), on the axes immediately below. View the scale of the grid for the graph of \(f\) as being \(1 \times 1\), and assume the horizontal scale of the grid for the graph of \(f'\) is identical to that for \(f\). If you need to adjust the vertical scale on the axes for the graph of \(f'\), you should label that accordingly.

Vis svaret

At any point where there is a jump in the graph of the derivative, the derivative is undefined. Normally we would draw an open circle at each end of the graph, but those are omitted here for convenience of plotting.
Let \(g\) be a continuous function (that is, one with no jumps or holes in the graph) and suppose that a graph of \(y= g'(x)\) is given by the graph on the right in Figure.
1. Observe that for every value of \(x\) that satisfies \(0 \lt x \lt 2\), the value of \(g'(x)\) is constant. What does this tell you about the behavior of the graph of \(y = g(x)\) on this interval?
2. On what intervals other than \(0 \lt x \lt 2\) do you expect \(y = g(x)\) to be a linear function? Why?
3. At which values of \(x\) is \(g'(x)\) not defined? What behavior does this lead you to expect to see in the graph of \(y=g(x)\)?
4. Suppose that \(g(0) = 1\). On the axes provided at left in Figure, sketch an accurate graph of \(y = g(x)\).
Vis svaret
1. Since \(g'(x)\) is constant (with value \(1\)) on the interval \(0 \lt x \lt 2\), it follows that \(g\) is linear on that same interval, since \(g\) is increasing at a constant rate.
2. On \(-3.5 \lt x \lt -2\), we also expect \(g\) to be linear with slope \(1\), while on \(-2 \lt x \lt 0\) and \(2 \lt x \lt 3.5\), \(g\) will be linear with slope \(-1\); in each case this is true because the value of \(g'(x)\) is constant with the noted value on the interval.
3. From the given graph of \(g'(x)\), we observe that \(g'\) is undefined at \(x = -2, 0, 2\). Since we have been given that \(g\) is a continuous function, we can conclude that \(g\) must have sharp corners on its graph at these points. Moreover, that makes sense in light of our earlier observations that show \(g\) has constant slope on the intervals that connect at \(x = -2, 0, 2\) and the graph jumps from having slope \(1\) to \(-1\), and then back, and so on.

Symbols used here

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: The derivative function

How does the limit definition of the derivative of a function f lead to an entirely new (but related) function f'?
What is the difference between writing f'(a) and f'(x)?
How is the graph of the derivative function f'(x) related to the graph of f(x)?
What are some examples of functions f for which f' is not defined at one or more points?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

Prøv din egen

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

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Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests