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The limit definition and what a derivative means with units; where a derivative fails to exist; the power, product and quotient rules and the five derivatives to know outright.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you can compute a derivative from the limit definition, read one as a rate with units, estimate one from a table, and say why a corner has no derivative even where the function is continuous. You can also differentiate with the power, product and quotient rules, and know the derivatives of $\sin$, $\cos$, $\tan$, $e^x$ and $\ln x$ without deriving them.
You can find the slope of a straight line, and you have just learned to evaluate a limit of the form $\tfrac{0}{0}$. A derivative is exactly those two things put together: the slope of a line through two points, as the two points close up on one.
Average rate of change: $\dfrac{f(b) - f(a)}{b - a}$, the slope of the secant through two points.
Instantaneous rate of change: the limit of that as the two points merge — the derivative.
Tangent line: the line through one point with the derivative as its slope.
Differentiable: the limit exists. Corners, cusps and vertical tangents are the ways it fails.
A straight line has one slope everywhere, and a curve does not — so 'the slope of a curve at a point' has to be defined rather than measured. The definition is a limit:
$$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$
The fraction inside is the slope of the secant through $(a, f(a))$ and a nearby point. As $h$ shrinks the second point slides toward the first, the secants pivot, and their slopes settle on a number — the slope of the tangent, and the function's instantaneous rate of change.
Both readings matter and they are the same fact. Geometrically $f'(a)$ is a slope; in a context it is a rate, with the units of the output over the units of the input. A population's derivative is people per year; a cost's is pounds per item; a position's is metres per second, which is a velocity.
A derivative can fail to exist, and there are exactly three ways: a corner, where the two sides give different slopes; a cusp, where they run off in opposite directions; and a vertical tangent, where the slope is unbounded. All three can happen at a point where the function is perfectly continuous — so differentiability implies continuity, and the converse is false.
Another way: picture
Picture a curve with a fixed point on it and a second point sliding toward it. Draw the line through both. As the sliding point gets close, the line stops swinging and settles into one position: the tangent. The derivative is that final slope, and the whole of the definition is the arithmetic of the sliding.
Another way: steps
"$\dfrac{f(a+h) - f(a)}{h}$ is $\tfrac{0}{0}$, so it is meaningless." It is $\tfrac{0}{0}$ in the limit, which is why the algebra comes first: expand, cancel the $h$, and the expression that remains has a perfectly ordinary limit.
"Continuous means differentiable." $|x|$ is continuous everywhere and has no derivative at zero. Differentiability is strictly stronger: it requires the slope to settle down, not merely the value.
"The derivative is a value of the function." It is a rate, and it has different units — litres per minute, not litres. Nothing goes wrong more quietly than reading a derivative as a quantity.
$f(2 + h) = 3(2 + h)^2 = 12 + 12h + 3h^2$.
Expand before subtracting.
Subtract $f(2) = 12$: what is left is $12h + 3h^2$.
The constants go.
Divide by $h$: $12 + 3h$. Let $h \to 0$: $f'(2) = 12$.
$C(x)$ is the cost in pounds of making $x$ chairs, and $C'(50) = 34$.
Given.
The units are pounds per chair, so making the fifty-first chair costs about £34.
Output units over input units.
It is not the cost of fifty chairs, and it is not the average cost per chair. It is the cost of one more.
$f(1 + h) = 1 + 2h + h^2 + 5 + 5h = 6 + 7h + h^2$.
Subtract $f(1) = 6$ and divide by $h$: $7 + h$.
Let $h \to 0$: $f'(1) = 7$ — which the power rule will give in one line.
For $f(x) = 9x^2$, use $\displaystyle\lim_{h \to 0} \frac{f(5 + h) - f(5)}{h}$ to find $f'(5)$.
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A tank holds $V(t)$ litres after $t$ minutes, and $V'(3) = 6$. What does that say?
A table gives $f(4) = 24$ and $f(6) = 38$, and nothing between. Estimate $f'(4)$.
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A reservoir holds $25$ million litres and is falling at $8$ million litres a week, a rate that is itself easing. Using only the rate at this instant, estimate the volume in $4$ weeks — and say whether the estimate is too high or too low.
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You can compute a derivative from the definition. Doing that for every function would be unbearable, so this section replaces it with rules — each of which was proved from the definition once, so that nobody has to do it again.
Power rule: $\dfrac{d}{dx}x^n = nx^{n-1}$.
Linearity: the derivative of a sum is the sum of the derivatives, and constants come out front.
Product rule: $(fg)' = f'g + fg'$.
Quotient rule: $\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}$.
Every rule below was proved from the limit definition once and for all.
Power and linearity. $\dfrac{d}{dx}x^n = nx^{n-1}$ for every $n$, including negative and fractional ones. Derivatives of sums are sums of derivatives, and a constant multiple comes straight out front — so a polynomial is differentiated term by term, and a constant term vanishes.
Product. $(fg)' = f'g + fg'$. Two terms, because a product can change for two reasons: the first factor moves, or the second does.
Quotient. $\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}$. The order in the numerator is not negotiable.
The five to know outright.
$$\frac{d}{dx}\sin x = \cos x, \quad \frac{d}{dx}\cos x = -\sin x, \quad \frac{d}{dx}\tan x = \sec^2 x$$ $$\frac{d}{dx}e^x = e^x, \quad \frac{d}{dx}\ln x = \frac{1}{x}$$
Two of those are worth pausing on. $e^x$ is its own derivative — it is the function whose rate of growth equals its size, which is what singles out $e$ among all the bases. And $\ln x$, a transcendental function, has an algebraic derivative, which is why logarithms turn up in the answers to integrals that contain no logarithm at all.
Another way: picture
Picture a rectangle whose width is $f$ and whose height is $g$, so its area is $fg$. Nudge both a little. The area grows by a strip along the top, $f \, dg$, plus a strip along the side, $g \, df$, plus a tiny corner square too small to matter. That is the product rule, and the corner square is why there is no third term.
Another way: steps
"$(fg)' = f'g'$." It is not, and the quickest disproof is $x \cdot x$: the left side is $2x$ and the wrong right side is $1$. The product rule has two terms because a product changes for two reasons at once.
"The quotient rule is symmetric." It is not: $f'g - fg'$, in that order, and getting it backwards negates the whole answer. If you cannot remember which way round, derive it from the product rule.
"$\dfrac{d}{dx}\cos x = \sin x$." It is $-\sin x$. The minus belongs to the cosine, and this single sign accounts for more lost marks than any other fact in the course.
$f(x) = 3x^2(2x + 1)$, so $u = 3x^2$ and $v = 2x + 1$.
Name the parts.
$u' = 6x$ and $v' = 2$.
$u'v + uv' = 6x(2x+1) + 3x^2(2) = 12x^2 + 6x + 6x^2 = 18x^2 + 6x$.
Two terms, then tidy.
$f(x) = \dfrac{5x}{x + 2}$: $u = 5x$, $v = x + 2$.
$\dfrac{5(x + 2) - 5x(1)}{(x+2)^2} = \dfrac{10}{(x+2)^2}$.
The $x$ terms cancel.
The derivative is positive everywhere, which matches the graph: the function climbs toward its asymptote and never turns back.
A product: $u = x^2$, $v = \sin x$.
Name the structure first.
$u' = 2x$ and $v' = \cos x$.
$f'(x) = 2x\sin x + x^2\cos x$ — two terms, and no simplification helps.
Differentiate $f(x) = 4x^3 - 8x^2 + 3x$.
Answer:
Differentiate $f(x) = 5x^2(4x + 3)$, and give the result expanded.
Answer:
For $f(x) = \dfrac{2x}{x + 5}$, find $f'(4)$.
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Match each function to its derivative.
| $\cos x$ | $-\sin x$ | $e^x$ | $\dfrac{1}{x}$ | |
|---|---|---|---|---|
| $\sin x$ | ||||
| $\cos x$ | ||||
| $e^x$ | ||||
| $\ln x$ |
For $f(x) = 2x^3 + 5x$, find the slope of the tangent at $x = 3$.
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Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Why does $f(x) = |x - 5|$ have no derivative at $x = 5$?
What is the derivative of $\sin x$?
You can differentiate from the definition and with the rules. Without looking: what does $f'(a)$ mean geometrically and in units, and why is $(fg)'$ not $f'g'$?
8. Your turn: $f(x) = x^2 + 5x$ at $x = 1$, step 3
19. Your turn: $f(x) = x^2 \sin x$, step 3