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Introduction to the derivative

3.3 The derivative

36 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

The function f(x)f(x) is shown. Select all options below that describe its derivative, dfdx\ds\diff{f}{x}:

(a) constant (b) increasing (c) decreasing
(d) always positive (e) always negative

Figure from prob_s2.2, line 7

Figure from prob_s2.2, line 7

Hint

What are the properties of f′f' when ff is a line?

Answer

(a), (d)

Full solution
Q2Stage 1

The function f(x)f(x) is shown. Select all options below that describe its derivative, dfdx\ds\diff{f}{x}:

(a) constant (b) increasing (c) decreasing
(d) always positive (e) always negative

Figure from prob_s2.2, line 7

Figure from prob_s2.2, line 7

Hint

Be very careful not to confuse ff and f′f'.

Answer

(e)

Full solution
Q3Stage 1

The function f(x)f(x) is shown. Select all options below that describe its derivative, dfdx\ds\diff{f}{x}:

(a) constant (b) increasing (c) decreasing
(d) always positive (e) always negative

Figure from prob_s2.2, line 6

Figure from prob_s2.2, line 6

Hint

Be very careful not to confuse ff and f′f'.

Answer

(b)

Full solution
Q4Stage 1Past exam · 2006H

State, in terms of a limit, what it means for f(x)=x3f(x) = x^3 to be differentiable at x=0x = 0.

Answer

By definition, f(x)=x3f(x) = x^3 is differentiable at x=0x = 0 if the limit

lim⁡h→0f(h)−f(0)h=lim⁡h→0h3−0h\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h} =\lim_{h\rightarrow 0}\frac{h^3-0}{h}

exists.

Full solution
Q5Stage 1

For which values of xx does f′(x)f'(x) not exist?

Figure from prob_s2.2, line 2

Figure from prob_s2.2, line 2

Hint

The slope has to look “the same" from the left and the right.

Answer

x=−1x=-1 and x=3x=3

Full solution
Q6Stage 1

Suppose f(x)f(x) is a function defined at x=ax=a with

lim⁡h→0−f(a+h)−f(a)h=lim⁡h→0+f(a+h)−f(a)h=1.\lim_{h \to 0^-}\frac{f(a+h)-f(a)}{h}=\lim_{h \to 0^+}\frac{f(a+h)-f(a)}{h}=1.

True or false: f′(a)=1f'(a)=1.

Hint

Use the definition of the derivative, and what you know about limits.

Answer

True. (Contrast to Question 7.)

Full solution
Q7Stage 1

Suppose f(x)f(x) is a function defined at x=ax=a with

lim⁡x→a−f′(x)=lim⁡x→a+f′(x)=1.\lim_{x \to a^-}f'(x)=\lim_{x \to a^+}f'(x)=1.

True or false: f′(a)=1f'(a)=1.

Hint

Consider continuity.

Answer

In general, false. (Contrast to Question 6.)

Full solution
Q8Stage 1

Suppose s(t)s(t) is a function, with tt measured in seconds, and ss measured in metres. What are the units of s′(t)s'(t)?

Hint

Look at the definition of the derivative. Your answer will be a fraction.

Answer

metres per second

Full solution

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q9Stage 2

Use the definition of the derivative to find the equation of the tangent line to the curve y(x)=x3+5y(x)=x^3+5 at the point (1,6)(1,6).

Hint

You need a point (given), and a slope (derivative).

Answer

y−6=3(x−1)y-6=3(x-1), or y=3x+3y=3x +3

Full solution
Q10Stage 2

Use the definition of the derivative to find the derivative of f(x)=1xf(x)=\frac{1}{x}.

Hint

You'll need to add some fractions.

Answer

−1x2\dfrac{-1}{x^2}

Full solution
Q11Stage 2Past exam · 2007H

Let f(x)=x∣x∣f(x) = x|x|. Using the definition of the derivative, show that f(x)f(x) is differentiable at x=0x = 0.

Hint

You don't have to take the limit from the left and right separately–things will cancel nicely.

Answer

By definition

f′(0)=lim⁡h→0f(h)−f(0)h=lim⁡h→0h∣h∣h=lim⁡h→0∣h∣=0f'(0)=\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h} =\lim_{h\rightarrow 0}\frac{h|h|}{h} =\lim_{h\rightarrow 0}|h|=0

In particular, the limit exists, so the derivative exists (and is equal to zero).

Full solution
Q12Stage 2Past exam · 1997A

Use the definition of the derivative to compute the derivative of the function f(x)=2x+1f(x)=\frac{2}{x+1}.

Hint

You might have to add fractions

Answer

−2(x+1)2\dfrac{-2}{(x+1)^2}

Full solution
Q13Stage 2Past exam · 1996D

Use the definition of the derivative to compute the derivative of the function f(x)=1x2+3f(x)=\frac{1}{x^2+3}.

Answer

−2x[x2+3]2\dfrac{-2x}{[x^2+3]^2}

Full solution
Q14Stage 2

Use the definition of the derivative to find the slope of the tangent line to the curve
f(x)=xlog⁡10(2x+10)f(x)=x\log_{10}(2x+10) at the point x=0x=0.

Hint

Your limit should be easy.

Answer

1

Full solution
Q15Stage 2Past exam · 1998H

Compute the derivative of f(x)=1x2f(x)=\frac{1}{x^2} directly from the definition.

Hint

add fractions

Answer

f′(x)=−2x3f'(x)=-\dfrac{2}{x^3}

Full solution
Q16Stage 2Past exam · 2006H

Find the values of the constants aa and bb for which

f(x)={x2x≤2ax+bx>2\begin{align*} f(x) = \left\{ \begin{array}{lc} x^2 & x\le 2\\ ax + b & x > 2 \end{array}\right. \end{align*}

is differentiable everywhere.

Remark: In the text, you have already learned the derivatives of x2x^2 and ax+bax+b. In this question, you are only asked to find the values of aa and bb—not to justify how you got them—so you don't have to use the definition of the derivative. However, on an exam, you might be asked to justify your answer, in which case you would show how to differentiate the two branches of f(x)f(x) using the definition of a derivative.

Hint

For ff to be differentiable at x=2x=2, two things must be true: it must be continuous at x=2x=2, and the derivative from the right must equal the derivative from the left.

Answer

a=4a=4, b=−4b=-4

Full solution
Q17Stage 2Past exam · 2009H

Use the definition of the derivative to compute f′(x)f'(x) if f(x)=1+xf(x) = \sqrt{1 + x}. Where does f′(x)f'(x) exist?

Hint

After you plug in f(x)f(x) to the definition of a derivative, you'll want to multiply and divide by the conjugate 1+x+h+1+x\sqrt{1+x+h}+\sqrt{1+x}.

Answer

f′(x)=121+xf'(x)=\dfrac{1}{2\sqrt{1+x}} when x>−1x>-1; f′(x)f'(x) does not exist when x≤−1x \leq -1.

Full solution

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q18Stage 3

Use the definition of the derivative to find the velocity of an object whose position is given by the function s(t)=t4−t2s(t)=t^4-t^2.

Hint

When you're finding the derivative, you'll need to cancel a lot on the numerator, which you can do by expanding the polynomials.

Answer

v(t)=4t3−2tv(t)=4t^3-2t

Full solution
Q19Stage 3Past exam · 2015Q

Determine whether the derivative of following function exists at x=0x=0.

f(x)={xcos⁡x if x≥0x2+x4 if x<0\begin{align*} f(x) &=\begin{cases} x \cos x & \text{ if } x\ge 0\\ \sqrt{x^2+x^4} & \text{ if } x< 0 \end{cases} \end{align*}

You must justify your answer using the definition of a derivative.

Hint

You'll need to look at limits from the left and right. The fact that f(0)=0f(0)=0 is useful for your computation. Recall that if x<0x<0 then x2=∣x∣=−x\sqrt{x^2}=|x|=-x.

Answer

No, it does not.

Full solution
Q20Stage 3Past exam · 2015Q

Determine whether the derivative of the following function exists at x=0x=0

f(x)={xcos⁡x if x≤01+x−1 if x>0\begin{align*} f(x) &=\begin{cases} x \cos x & \text{ if } x\le 0\\ \sqrt{1+x}-1 & \text{ if } x> 0 \end{cases} \end{align*}

You must justify your answer using the definition of a derivative.

Hint

You'll need to look at limits from the left and right. The fact that f(0)=0f(0)=0 is useful for your computation.

Answer

No, it does not.

Full solution
Q21Stage 3Past exam · 2015Q

Determine whether the derivative of the following function exists at x=0x=0

f(x)={x3−7x2 if x≤0x3cos⁡(1x) if x>0\begin{align*} f(x) &=\begin{cases} x^3-7x^2 & \text{ if } x\le 0\\ x^3 \cos\left(\frac{1}{x}\right) & \text{ if } x> 0 \end{cases} \end{align*}

You must justify your answer using the definition of a derivative.

Hint

You'll need to look at limits from the left and right. The fact that f(0)=0f(0)=0 is useful for your computation.

Answer

Yes, it is.

Full solution
Q22Stage 3Past exam · 2015Q

Determine whether the derivative of the following function exists at x=1x=1

f(x)={4x2−8x+4 if x≤1(x−1)2sin⁡(1x−1) if x>1\begin{align*} f(x) &=\begin{cases} 4x^2-8x+4 & \text{ if } x\le 1\\ (x-1)^2\sin\left(\dfrac{1}{x-1}\right) & \text{ if } x> 1 \end{cases} \end{align*}

You must justify your answer using the definition of a derivative.

Hint

You'll need to look at limits from the left and right. The fact that f(1)=0f(1)=0 is useful for your computation.

Answer

Yes, it is.

Full solution
Q23Stage 3

Sketch a function f(x)f(x) with f′(0)=−1f'(0)=-1 that takes the following values:

x\mathbf{x}−1-1−12-\frac{1^{ }}{2_{ }}−14-\frac{1}{4}−18-\frac{1}{8}0018\frac{1}{8}14\frac{1}{4}12\frac{1}{2}11
f(x)\mathbf{f(x)}−1-1−12-\frac{1^{ }}{2_{ }}−14-\frac{1}{4}−18-\frac{1}{8}0018\frac{1}{8}14\frac{1}{4}12\frac{1}{2}11

Remark: you can't always guess the behaviour of a function from its points, even if the points seem to be making a clear pattern.

Hint

There's lots of room between 00 and 18\frac{1}{8}; see what you can do with it.

Answer

Many answers are possible; here is one.

Figure from prob_s2.2, line 2

Figure from prob_s2.2, line 2

Full solution
Q24Stage 3

Let p(x)=f(x)+g(x)p(x)=f(x)+g(x), for some functions ff and gg whose derivatives exist. Use limit laws and the definition of a derivative to show that p′(x)=f′(x)+g′(x)p'(x)=f'(x)+g'(x).

Remark: this is called the sum rule, and we'll learn more about it in Lemma 4.1.1.

Hint

Set up your usual limit, then split it into two pieces

Answer
p′(x)=lim⁡h→0p(x+h)−p(x)h=lim⁡h→0f(x+h)+g(x+h)−f(x)−g(x)h=lim⁡h→0f(x+h)−f(x)+g(x+h)−g(x)h=lim⁡h→0[f(x+h)−f(x)h+g(x+h)−g(x)h](∗)=[lim⁡h→0f(x+h)−f(x)h]+[lim⁡h→0g(x+h)−g(x)h]=f′(x)+g′(x)\begin{align*} p'(x) &= \lim_{h \rightarrow 0} \frac{p(x+h)-p(x)}{h}\\ &= \lim_{h \rightarrow 0} \frac{f(x+h)+g(x+h)-f(x)-g(x)}{h}\\ &= \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)+g(x+h)-g(x)}{h}\\ &= \lim_{h \rightarrow 0} \left[\frac{f(x+h)-f(x)}{h}+ \frac{g(x+h)-g(x)}{h}\right]\\ (*)&= \left[\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right]+ \left[\lim_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}\right]\\ &= f'(x)+g'(x) \end{align*}

At step (∗*), we use the limit law that lim⁡x→a[F(x)+G(x)]=lim⁡x→aF(x)+lim⁡x→aG(x)\displaystyle\lim_{x \rightarrow a} \left[F(x)+G(x)\right] = \displaystyle\lim_{x \rightarrow a} F(x)+\displaystyle\lim_{x \rightarrow a}G(x), as long as lim⁡x→aF(x)\displaystyle\lim_{x \rightarrow a} F(x) and lim⁡x→aG(x)\displaystyle\lim_{x \rightarrow a}G(x) exist. Because the problem states that f′(x)f'(x) and g′(x)g'(x) exist, we know that lim⁡h→0f(x+h)−f(x)h\displaystyle\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} and lim⁡h→0g(x+h)−g(x)h\displaystyle\lim_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h} exist, so our work is valid.

Full solution
Q25Stage 3

Let f(x)=2xf(x)=2x, g(x)=xg(x)=x, and p(x)=f(x)⋅g(x)p(x)=f(x) \cdot g(x).

  1. Find f′(x)f'(x) and g′(x)g'(x).

  2. Find p′(x)p'(x).

  3. Is p′(x)=f′(x)⋅g′(x)p'(x)=f'(x) \cdot g'(x)?

In Theorem 4.1.3, you'll learn a rule for calculating the derivative of a product of two functions.

Hint

You don't need the definition of the derivative for a line.

Answer

(a) f′(x)=2f'(x)=2 and g′(x)=1g'(x)=1 (b) p′(x)=4xp'(x)=4x (c) no

Full solution
Q26Stage 3Past exam · 2006H

There are two distinct straight lines that pass through the point (1,−3)(1,-3) and are tangent to the curve y=x2y = x^2. Find equations for these two lines.

Remark: the point (1,−3)(1,-3) does not lie on the curve y=x2y=x^2.

Hint

A generic point on the curve has coordinates (α,α2)(\alpha, \alpha^2). In terms of α\alpha, what is the equation of the tangent line to the curve at the point (α,α2)(\alpha, \alpha^2)? What does it mean for (1,−3)(1,-3) to be on that line?

Answer

y=6x−9y=6x-9 and y=−2x−1y=-2x-1

Full solution
Q27Stage 3Past exam · 2009H

For which values of aa is the function

f(x)={0x≤0xasin⁡1xx>0f(x) =\left\{\begin{array}{ll} 0 & x\le 0\\ x^a \sin\frac{1}{x} & x > 0\end{array}\right.

differentiable at 0?

Hint

Remember for a constant nn,

lim⁡h→0hn={0n>01n=0DNEn<0\ds\lim_{h \to 0} h^{n} = \left\{\begin{array}{ll} 0&n>0\\ 1&n=0\\ DNE&n<0 \end{array}\right.
Answer

a>1a>1

Full solution

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q28Stage 2

Suppose h(t)h(t) gives the height at time tt of the water at a dam, where the units of tt are hours and the units of hh are meters.

  1. What is the physical interpretation of the slope of the secant line through the points (0,h(0))(0,h(0)) and (24,h(24))(24,h(24))?

  2. What is the physical interpretation of the slope of the tangent line to the curve y=h(t)y=h(t) at the point (0,h(0))(0,h(0))?

Hint

Think about units.

Answer

(a) The average rate of change of the height of the water over the single day starting at t=0t=0, measured in mhr\frac{\mathrm{m}}{\mathrm{hr}}.

(b) The instantaneous rate of change of the height of the water at the time t=0t=0.

Full solution
Q29Stage 2

Suppose p(t)p(t) is a function that gives the profit generated by selling tt widgets. What is the practical interpretation of p′(t)p'(t)?

Answer

Profit per additional widget sold, when tt widgets are being sold. This is called the marginal profit per widget, when tt widgets are being sold.

Full solution
Q30Stage 2

T(d)T(d) gives the temperature of water at a particular location dd metres below the surface. What is the physical interpretation of T′(d)T'(d)? Would you expect the magnitude of T′(d)T'(d) to be larger when dd is near 0, or when dd is very large?

Answer

T′(d)T'(d) measures how quickly the temperature is changing per unit change of depth, measured in degrees per metre. ∣T′(d)∣|T'(d)| will probably be largest when dd is near zero, unless there are hot springs or other underwater heat sources.

Full solution
Q31Stage 2

C(w)C(w) gives the calories in ww grams of a particular dish. What does C′(w)C'(w) describe?

Answer

Calories per additional gram, when there are ww grams.

Full solution
Q32Stage 2

The velocity of a moving object at time tt is given by v(t)v(t). What is v′(t)v'(t)?

Answer

The acceleration of the object.

Full solution
Q33Stage 2

The function T(j)T(j) gives the temperature in degrees Celsius of a cup of water after jj joules of heat have been added. What is T′(j)T'(j)?

Answer

Degrees Celsius temperature change per joule of heat added. (This is closely related to heat capacity and to specific heat — there's a nice explanation of this on Wikipedia.)

Full solution
Q34Stage 2

A population of bacteria, left for a fixed amount of time at temperature TT, grows to P(T)P(T) individuals. Interpret P′(T)P'(T).

Answer

Number of bacteria added per degree. That is: the number of extra bacteria (possibly negative) that will exist in the population by raising the temperature by one degree.

Full solution

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q35Stage 3

You hammer a small nail into a wooden wagon wheel. R(t)R(t) gives the number of rotations the nail has undergone tt seconds after the wagon started to roll. Give an equation for how quickly the nail is rotating, measured in degrees per second.

Hint

There are 360 degrees in one rotation.

Answer

360R′(t)360R'(t)

Full solution
Q36Stage 3

A population of bacteria, left for a fixed amount of time at temperature TT, grows to P(T)P(T) individuals. There is one ideal temperature where the bacteria population grows largest, and the closer the sample is to that temperature, the larger the population is (unless the temperature is so extreme that it causes all the bacteria to die by freezing or boiling). How will P′(T)P'(T) tell you whether you are colder or hotter than the ideal temperature?

Hint

P′(t)P'(t) was discussed in Question 34.

Answer

If P′(t)P'(t) is positive, your sample is below the ideal temperature, and if P′(t)P'(t) is negative, your sample is above the ideal temperature. If P′(t)=0P'(t) = 0, you don't know whether the sample is exactly at the ideal temperature, or way above or below it with no living bacteria.

Full solution

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From the UBC Math 100 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.