Module 41 of 47Lesson 1 of 10 · HL · AI 5.9 — Further differentiation: chain, product, quotient rules and related rates

Course map · Topic 5: Calculus

5.9 Further differentiation: chain, product, quotient rules and related rates

HL only11 worked questions

At HL you must differentiate much more than powers of x: sin, cos, tan, eˣ and ln x, and functions built from them by multiplying, dividing or putting one inside another. Related rates then use the chain rule to link how fast two connected quantities change.

✏️ Start with the questions. Try each one on paper, then open the solution and compare.

Stuck? Open the notes ↓

✏️ Practice questions

Try each question on paper first. Then open the solution and compare every step.

Question 1●○○ Warm-up✍️ By hand · Paper 1[5 marks]

Differentiate each function with respect to

x

(a) $y=3\sin x-2\cos x+4\mathrm{e}^{x}$ [2]
(b) $f(x)=4\sqrt{x}-\dfrac{3}{x^{2}}+2\ln x$ . Hence find the gradient of the graph of $f$ at $x=1$ .[3]

Show the full step-by-step solution

Step 1🧠 ThinkUse the list of standard derivatives

Each term is on its own (added or subtracted), so we differentiate one term at a time. A number in front just stays in front. Learn the list: sin becomes cos, cos becomes minus sin,

\mathrm{e}^x

stays

\mathrm{e}^x

\ln x

becomes

\dfrac{1}{x}

Part (a)

Step 1✍️ By handOne term at a time

3\sin x \to 3\cos x

-2\cos x \to -2\times(-\sin x)=+2\sin x

4\mathrm{e}^x\to4\mathrm{e}^x

\dfrac{\mathrm{d}y}{\mathrm{d}x}=3\cos x+2\sin x+4\mathrm{e}^{x}

Part (b)

Step 1🧠 ThinkFirst rewrite as powers of x

The power rule needs powers. A square root is the power

\dfrac{1}{2}

. Dividing by

x^2

is the power

-2

Step 2✍️ By handRewrite, then differentiate

f(x)=4x^{\frac{1}{2}}-3x^{-2}+2\ln x

f'(x)=4\times\dfrac{1}{2}x^{-\frac{1}{2}}-3\times(-2)x^{-3}+2\times\dfrac{1}{x}

f'(x)=\dfrac{2}{\sqrt{x}}+\dfrac{6}{x^{3}}+\dfrac{2}{x}

f'(1)=2+6+2=10

✅ Check it

-3\times(-2)=+6

: two minus signs make a plus. That is the step most students get wrong.

Final answer (a)

\dfrac{\mathrm{d}y}{\mathrm{d}x}=3\cos x+2\sin x+4\mathrm{e}^{x}

(b)

f'(x)=\dfrac{2}{\sqrt{x}}+\dfrac{6}{x^{3}}+\dfrac{2}{x}

; gradient at

x=1

10

⚠️ Common mistake Writing the derivative of

\cos x

+\sin x

. It is

-\sin x

Question 2●○○ Warm-up✍️ By hand · Paper 1[7 marks]

Use the chain rule to differentiate:

(a) $y=(3x^{2}-5)^{4}$ [2]
(b) $y=\mathrm{e}^{2x+1}$ [1]
(c) $y=\sin(5x)$ [1]
(d) $y=\dfrac{1}{\sqrt{4x+1}}$ [3]

Show the full step-by-step solution

Step 1🧠 ThinkSpot the “inside” and the “outside”

Each of these is a function inside another function. In

(3x^2-5)^4

the inside is

3x^2-5

and the outside is “to the power 4”. The chain rule says: differentiate the outside (leave the inside alone), then multiply by the derivative of the inside.

Part (a)

Step 1✍️ By handOutside: power 4. Inside: 3x² − 5

Let

u=3x^2-5

, so

y=u^4

and

\dfrac{\mathrm{d}u}{\mathrm{d}x}=6x

\dfrac{\mathrm{d}y}{\mathrm{d}x}=4u^{3}\times6x

\dfrac{\mathrm{d}y}{\mathrm{d}x}=24x(3x^{2}-5)^{3}

Part (b)

Step 1✍️ By hande to a power keeps itself, times the derivative of the power

\dfrac{\mathrm{d}y}{\mathrm{d}x}=\mathrm{e}^{2x+1}\times2=2\mathrm{e}^{2x+1}

Part (c)

Step 1✍️ By handsin becomes cos, times the derivative of 5x

\dfrac{\mathrm{d}y}{\mathrm{d}x}=\cos(5x)\times5=5\cos(5x)

Part (d)

Step 1✍️ By handRewrite as a power first

y=(4x+1)^{-\frac{1}{2}}

\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{1}{2}(4x+1)^{-\frac{3}{2}}\times4

M1A1

\dfrac{\mathrm{d}y}{\mathrm{d}x}=-2(4x+1)^{-\frac{3}{2}}=-\dfrac{2}{(4x+1)^{\frac{3}{2}}}

💡 Remember Quick pattern for a linear inside

ax+b

: just differentiate as normal and multiply by

a

. E.g.

\dfrac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{7x}=7\mathrm{e}^{7x}

Final answer (a)

24x(3x^{2}-5)^{3}

(b)

2\mathrm{e}^{2x+1}

(c)

5\cos(5x)

(d)

-\dfrac{2}{(4x+1)^{\frac{3}{2}}}

⚠️ Common mistake In (a), writing

4(3x^2-5)^3

and forgetting the

\times6x

from the inside.

Question 3●○○ Warm-up✍️ + 🖩 · Paper 1[4 marks]

Let

f(x)=x^{3}\mathrm{e}^{2x}

(a) Find $f'(x)$ .[3]
(b) Find the exact gradient of the graph of $f$ at $x=\dfrac{1}{2}$ .[1]

Show the full step-by-step solution

Step 1🧠 ThinkTwo functions multiplied → product rule

x^3

and

\mathrm{e}^{2x}

are multiplied together. Call the first one

u

and the second one

v

. The product rule says

f'=uv'+vu'

: “first times derivative of the second, plus second times derivative of the first”.

Part (a)

Step 1✍️ By handList u, v and their derivatives first

u=x^{3}\qquad u'=3x^{2}

v=\mathrm{e}^{2x}\qquad v'=2\mathrm{e}^{2x}

(chain rule)

f'(x)=x^{3}\times2\mathrm{e}^{2x}+\mathrm{e}^{2x}\times3x^{2}

M1A1

f'(x)=x^{2}\mathrm{e}^{2x}(2x+3)

Part (b)

Step 1✍️ By handSubstitute x = ½

f'\!\left(\dfrac{1}{2}\right)=\dfrac{1}{4}\mathrm{e}^{1}(1+3)=\mathrm{e}

Step 2🖩 On the GDCCheck with the numerical derivative

menu💻 click menu on keypad›4: Calculus›1: Numerical Derivative at a Point…›x›tab💻 Tab›0.5›enter💻 Enter›x^3·e^(2x)›enter💻 Enter

The dialog asks for the variable, the value and 1st or 2nd derivative. It pastes the d/dx template; type the function inside and press enter.

Both ≈ 2.718, so the exact answer e is right

Final answer (a)

f'(x)=2x^{3}\mathrm{e}^{2x}+3x^{2}\mathrm{e}^{2x}=x^{2}\mathrm{e}^{2x}(2x+3)

(b)

\mathrm{e}

⚠️ Common mistake Writing

f'(x)=3x^2\times2\mathrm{e}^{2x}

— just multiplying the two derivatives. That is not the product rule.

Question 4●●○ Standard✍️ + 🖩 · Paper 1[6 marks]

Let

f(x)=\dfrac{\ln x}{x}

, for

x\gt 0

(a) Show that $f'(x)=\dfrac{1-\ln x}{x^{2}}$ .[3]
(b) Hence find the exact coordinates of the maximum point of the graph of $f$ .[3]

Show the full step-by-step solution

Step 1🧠 ThinkOne function divided by another → quotient rule

Top

u=\ln x

, bottom

v=x

. The quotient rule:

f'=\dfrac{vu'-uv'}{v^2}

— “bottom times derivative of top, minus top times derivative of bottom, all over bottom squared”.

Part (a)

Step 1✍️ By handQuotient rule

u=\ln x,\ u'=\dfrac{1}{x}\qquad v=x,\ v'=1

f'(x)=\dfrac{x\times\dfrac{1}{x}-\ln x\times1}{x^{2}}

M1A1

f'(x)=\dfrac{1-\ln x}{x^{2}}

Part (b)

Step 1🧠 ThinkAt a maximum the gradient is 0

A fraction is 0 only when its top is 0 (the bottom

x^2

is never 0 here).

Step 2✍️ By handSolve f′(x) = 0

1-\ln x=0\Rightarrow\ln x=1\Rightarrow x=\mathrm{e}

M1A1

f(\mathrm{e})=\dfrac{\ln\mathrm{e}}{\mathrm{e}}=\dfrac{1}{\mathrm{e}}

\quad\Rightarrow\left(\mathrm{e},\ \dfrac{1}{\mathrm{e}}\right)

Step 3🖩 On the GDCCheck on a Graphs page

menu💻 click menu on keypad›6: Analyze Graph›3: Maximum

Graph f1(x) =

\dfrac{\ln (x)}{x}

, choose Maximum, then click to the left and to the right of the top of the curve.

Maximum at (2.718…, 0.3679…) = (e,

\dfrac{1}{e}

)

Final answer (a) shown (b)

\left(\mathrm{e},\ \dfrac{1}{\mathrm{e}}\right)\approx(2.72,\ 0.368)

⚠️ Common mistake Writing

uv'-vu'

on top. That gives the right size but the wrong sign.

Question 5●●○ Standard✍️ + 🖩 · Paper 1[5 marks]

The curve

y=3\sin(2x)

is defined for

0\le x\le\pi

. Find the equation of the tangent to the curve at the point where

x=\dfrac{\pi}{6}

. Give your answer in the form

y=mx+c

, with

c

correct to 3 significant figures.

Show the full step-by-step solution

Step 1🧠 ThinkA tangent needs a point and a gradient

The tangent is the straight line that just touches the curve. We need (1) the point on the curve, by putting

x=\dfrac{\pi}{6}

into

y

, and (2) the gradient, by putting

x=\dfrac{\pi}{6}

into

\dfrac{\mathrm{d}y}{\mathrm{d}x}

. The angle

\dfrac{\pi}{6}

is in radians.

Step 2✍️ By handGradient with the chain rule

\dfrac{\mathrm{d}y}{\mathrm{d}x}=3\cos(2x)\times2=6\cos(2x)

m=6\cos\left(\dfrac{\pi}{3}\right)=6\times\dfrac{1}{2}=3

Step 3✍️ By handThe point, then the line

y=3\sin\left(\dfrac{\pi}{3}\right)=\dfrac{3\sqrt3}{2}\approx2.598

y-2.598=3\left(x-\dfrac{\pi}{6}\right)

y=3x+1.03

Step 4🖩 On the GDCCheck: dy/dx on a Graphs page (RAD mode)

doc💻 File menu›menu💻 click menu on keypad›6: Analyze Graph›5: dy/dx

First set Angle to Radian (doc → Settings → Document Settings). Graph f1(x) = 3sin(2x), choose dy/dx, click the curve and type $\dfrac{\pi}{6}$ then enter. The calculator shows the gradient 3.

At x =

\dfrac{\pi}{6}

the gradient is 3

Final answer

y=3x+1.03

(exactly

y=3x+\dfrac{3\sqrt3}{2}-\dfrac{\pi}{2}

)

⚠️ Common mistake Calculator in DEG mode: it reads

\dfrac{\pi}{3}

as about 1.05 degrees, so the gradient shows about 0.105 instead of 3 and the point is wrong. Use RAD for all trig calculus.

Question 6●●○ Standard✍️ By hand · Paper 1[4 marks]

A stone is dropped into the still water of Gaborone Dam. A circular ripple spreads out so that its radius increases at a constant rate of 0.4 m s⁻¹. Find the rate at which the area inside the ripple is increasing when the radius is 5 m.

Show the full step-by-step solution

Step 1🧠 ThinkWrite down what you know and what you want

“Radius increases at 0.4 m per second” means

\dfrac{\mathrm{d}r}{\mathrm{d}t}=0.4

. We want how fast the area changes:

\dfrac{\mathrm{d}A}{\mathrm{d}t}

, at the moment when

r=5

. We need an equation linking

A

and

r

: the area of a circle.

Step 2✍️ By handLink, differentiate, then chain

A=\pi r^{2}\Rightarrow\dfrac{\mathrm{d}A}{\mathrm{d}r}=2\pi r

\dfrac{\mathrm{d}A}{\mathrm{d}t}=\dfrac{\mathrm{d}A}{\mathrm{d}r}\times\dfrac{\mathrm{d}r}{\mathrm{d}t}=2\pi r\times0.4

M1A1

When

r=5

\dfrac{\mathrm{d}A}{\mathrm{d}t}=2\pi\times5\times0.4=4\pi

\dfrac{\mathrm{d}A}{\mathrm{d}t}=4\pi\approx12.6\ \text{m}^2\,\text{s}^{-1}

💡 Check the units

\dfrac{\text{m}^2}{\text{m}}\times\dfrac{\text{m}}{\text{s}}=\dfrac{\text{m}^2}{\text{s}}

. If the units work out, the chain is set up correctly.

Final answer

4\pi\approx12.6\ \text{m}^2\,\text{s}^{-1}

⚠️ Common mistake Substituting

r=5

into

A=\pi r^2

first:

A=25\pi

, a constant, whose derivative is 0.

Question 7●●○ Standard✍️ By hand · Paper 1[4 marks]

Air is pumped into a spherical balloon at a rate of 50 cm³ s⁻¹. Find the rate at which the radius is increasing when the radius is 10 cm.

Show the full step-by-step solution

Step 1🧠 ThinkKnown rate and wanted rate

Volume is going up:

\dfrac{\mathrm{d}V}{\mathrm{d}t}=50

. We want

\dfrac{\mathrm{d}r}{\mathrm{d}t}

when

r=10

. The volume of a sphere

V=\dfrac{4}{3}\pi r^3

links them (it is in the formula booklet).

Step 2✍️ By handChain rule, then rearrange

\dfrac{\mathrm{d}V}{\mathrm{d}r}=4\pi r^{2}

\dfrac{\mathrm{d}V}{\mathrm{d}t}=\dfrac{\mathrm{d}V}{\mathrm{d}r}\times\dfrac{\mathrm{d}r}{\mathrm{d}t}\Rightarrow50=4\pi(10)^{2}\times\dfrac{\mathrm{d}r}{\mathrm{d}t}

M1A1

\dfrac{\mathrm{d}r}{\mathrm{d}t}=\dfrac{50}{400\pi}=\dfrac{1}{8\pi}\approx0.0398\ \text{cm s}^{-1}

✅ Check it Small answer? Yes: a balloon of radius 10 cm is already big (about 4 litres), so 50 cm³ more each second only grows the radius a little.

Final answer

\dfrac{1}{8\pi}\approx0.0398\ \text{cm s}^{-1}

Question 8●●● Exam-level✍️ By hand · Paper 1[8 marks]

A water tank at a farm near Francistown is in the shape of an inverted cone of top radius 2 m and height 5 m, as shown. Water is poured in at a rate of 0.3 m³ per minute. When the depth of the water is

h

m, the radius of the water surface is

r

(a) Show that the volume of water in the tank is $V=\dfrac{4\pi h^{3}}{75}$ .[2]
(b) Find the rate at which the depth is increasing when $h=3$ .[4]
(c) Find the rate at which the area of the water surface is increasing at this moment.[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkGet rid of r using similar triangles

The cone of water has the same shape as the whole tank. In the whole tank, radius : height = 2 : 5. So in the water,

\dfrac{r}{h}=\dfrac{2}{5}

, which means

r=\dfrac{2h}{5}

. Then the volume depends on

h

only.

Step 2✍️ By handSubstitute r = $\dfrac{2h}{5}$ into the cone formula

\dfrac{r}{h}=\dfrac{2}{5}\Rightarrow r=\dfrac{2h}{5}

V=\dfrac{1}{3}\pi r^{2}h=\dfrac{1}{3}\pi\left(\dfrac{2h}{5}\right)^{2}h=\dfrac{1}{3}\pi\times\dfrac{4h^{2}}{25}\times h=\dfrac{4\pi h^{3}}{75}

A1 AG

Part (b)

Step 1✍️ By handDifferentiate V with respect to h, then chain

\dfrac{\mathrm{d}V}{\mathrm{d}h}=\dfrac{12\pi h^{2}}{75}=\dfrac{4\pi h^{2}}{25}

\dfrac{\mathrm{d}V}{\mathrm{d}t}=\dfrac{\mathrm{d}V}{\mathrm{d}h}\times\dfrac{\mathrm{d}h}{\mathrm{d}t}\Rightarrow0.3=\dfrac{4\pi(3)^{2}}{25}\times\dfrac{\mathrm{d}h}{\mathrm{d}t}

M1A1

\dfrac{\mathrm{d}h}{\mathrm{d}t}=\dfrac{0.3\times25}{36\pi}=\dfrac{5}{24\pi}\approx0.0663\ \text{m per minute}

Part (c)

Step 1🧠 ThinkAnother linked quantity: the surface area

The water surface is a circle of radius

r=\dfrac{2h}{5}

, so

A=\pi r^2=\dfrac{4\pi h^{2}}{25}

. We already know

\dfrac{\mathrm{d}h}{\mathrm{d}t}

from (b).

Step 2✍️ By handChain again

\dfrac{\mathrm{d}A}{\mathrm{d}t}=\dfrac{\mathrm{d}A}{\mathrm{d}h}\times\dfrac{\mathrm{d}h}{\mathrm{d}t}=\dfrac{8\pi h}{25}\times\dfrac{5}{24\pi}

h=3

\dfrac{\mathrm{d}A}{\mathrm{d}t}=\dfrac{24\pi}{25}\times\dfrac{5}{24\pi}=0.2\ \text{m}^2\text{ per minute}

💡 Remember Always remove the extra variable (here

r

) before differentiating, so the formula has only the variable whose rate you need.

Final answer (a) shown (b)

\dfrac{5}{24\pi}\approx0.0663\ \text{m min}^{-1}

(c)

0.2\ \text{m}^2\ \text{min}^{-1}

⚠️ Common mistake Using

r=\dfrac{5h}{2}

(ratio upside down). Check: when

h=5

the radius must be 2.

Question 9●●● Exam-level✍️ By hand · Paper 1[7 marks]

A ladder of length 5 m leans against a vertical wall. The foot of the ladder slides away from the wall along horizontal ground at 0.5 m s⁻¹. Let

x

be the distance of the foot from the wall,

y

the height of the top of the ladder, and

\theta

the angle between the ladder and the ground.

(a) Find the rate at which the top of the ladder is moving down the wall when $x=3$ .[4]
(b) Find the rate of change of $\theta$ at this moment, in radians per second.[3]

Show the full step-by-step solution

Step 1🧠 ThinkWhat stays the same and what changes

The ladder's length (5 m) never changes;

x

y

and

\theta

all change with time. We know

\dfrac{\mathrm{d}x}{\mathrm{d}t}=0.5

(positive:

x

is getting bigger). Pythagoras links

x

and

y

Part (a)

Step 1✍️ By handDifferentiate Pythagoras with respect to time

Each term is differentiated with the chain rule:

\dfrac{\mathrm{d}}{\mathrm{d}t}(x^2)=2x\dfrac{\mathrm{d}x}{\mathrm{d}t}

x^{2}+y^{2}=25

2x\dfrac{\mathrm{d}x}{\mathrm{d}t}+2y\dfrac{\mathrm{d}y}{\mathrm{d}t}=0

M1A1

When

x=3

y=\sqrt{25-9}=4

2(3)(0.5)+2(4)\dfrac{\mathrm{d}y}{\mathrm{d}t}=0\Rightarrow\dfrac{\mathrm{d}y}{\mathrm{d}t}=-\dfrac{3}{8}=-0.375

The top moves down at

0.375\ \text{m s}^{-1}

Part (b)

Step 1✍️ By handLink θ and x with cosine

\cos\theta=\dfrac{x}{5}

-\sin\theta\,\dfrac{\mathrm{d}\theta}{\mathrm{d}t}=\dfrac{1}{5}\dfrac{\mathrm{d}x}{\mathrm{d}t}

M1A1

When

x=3

\sin\theta=\dfrac{y}{5}=\dfrac{4}{5}

-\dfrac{4}{5}\,\dfrac{\mathrm{d}\theta}{\mathrm{d}t}=\dfrac{0.5}{5}\Rightarrow\dfrac{\mathrm{d}\theta}{\mathrm{d}t}=-0.125\ \text{rad s}^{-1}

✅ Check it Both rates are negative: the top slides down and the ladder gets flatter. That matches the picture. ✔

Final answer (a)

\dfrac{\mathrm{d}y}{\mathrm{d}t}=-0.375

, i.e. down at

0.375\ \text{m s}^{-1}

(b)

\dfrac{\mathrm{d}\theta}{\mathrm{d}t}=-0.125\ \text{rad s}^{-1}

⚠️ Common mistake Putting

x=3

into

x^2+y^2=25

and then differentiating — you would lose

\dfrac{\mathrm{d}x}{\mathrm{d}t}

altogether.

Question 10●●● Exam-level✍️ + 🖩 · Paper 2[8 marks]

Mpho takes a dose of medicine. The concentration of the medicine in her blood, in mg per litre,

t

hours after the dose is modelled by

C(t)=20t\,\mathrm{e}^{-0.4t}

t\ge0

(a) Show that $C'(t)=20\mathrm{e}^{-0.4t}(1-0.4t)$ .[3]
(b) Find the time at which the concentration is greatest, and this greatest concentration.[3]
(c) Find the rate of change of the concentration 4 hours after the dose, and interpret its sign.[2]

Show the full step-by-step solution

Step 1🧠 ThinkProduct of two functions of t

20t

multiplied by

\mathrm{e}^{-0.4t}

→ product rule. And

\mathrm{e}^{-0.4t}

itself needs the chain rule (derivative of the power is

-0.4

Part (a)

Step 1✍️ By handProduct rule

u=20t,\ u'=20\qquad v=\mathrm{e}^{-0.4t},\ v'=-0.4\mathrm{e}^{-0.4t}

C'(t)=20t\times(-0.4\mathrm{e}^{-0.4t})+\mathrm{e}^{-0.4t}\times20

M1A1

C'(t)=20\mathrm{e}^{-0.4t}(1-0.4t)

A1 AG

Part (b)

Step 1✍️ By handGreatest value: C′(t) = 0

\mathrm{e}^{-0.4t}

is never zero, so the bracket must be zero.

1-0.4t=0\Rightarrow t=2.5

hoursM1A1

C(2.5)=50\mathrm{e}^{-1}\approx18.4

mg per litreA1

Step 2🖩 On the GDCConfirm with Maximum on a Graphs page

menu💻 click menu on keypad›6: Analyze Graph›3: Maximum

Maximum at (2.5, 18.394)

Part (c)

Step 1🖩 On the GDCGradient at t = 4

menu💻 click menu on keypad›4: Calculus›1: Numerical Derivative at a Point…

Use x for t on the calculator. Or substitute into (a):

20\mathrm{e}^{-1.6}(1-1.6)=-12\mathrm{e}^{-1.6}

C′(4) ≈ −2.42

Step 2✍️ By handInterpret

C'(4)\approx-2.42

mg per litre per hourA1

Negative: 4 hours after the dose the concentration is decreasing, by about 2.42 mg/L each hour.R1

Final answer (a) shown (b)

t=2.5

hours,

C=50\mathrm{e}^{-1}\approx18.4

mg/L (c)

-2.42

mg L⁻¹ h⁻¹; the concentration is falling.

Question 11●●○ Standard✍️ By hand · Paper 1[6 marks]

Differentiate with respect to

x

(a) $y=\ln(x^{2}+4)$ . Hence find the gradient of this curve at $x=2$ .[3]
(b) $y=\tan(3x)$ [1]
(c) $y=\mathrm{e}^{\sin x}$ [2]

Show the full step-by-step solution

Step 1🧠 ThinkAll three are chain-rule questions

In each one a whole expression sits inside ln, tan or e. Differentiate the outside function, keep the inside, and multiply by the derivative of the inside.

Part (a)

Step 1✍️ By handln of something → 1 over that something, times its derivative

\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{x^{2}+4}\times2x=\dfrac{2x}{x^{2}+4}

M1A1

x=2

\dfrac{4}{8}=\dfrac{1}{2}

Part (b)

Step 1✍️ By handtan → 1 over cos²

\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{\cos^{2}(3x)}\times3=\dfrac{3}{\cos^{2}(3x)}

Part (c)

Step 1✍️ By hande to a power: keep it, times the derivative of the power

\dfrac{\mathrm{d}y}{\mathrm{d}x}=\mathrm{e}^{\sin x}\times\cos x=\cos x\,\mathrm{e}^{\sin x}

M1A1

💡 Remember Useful shortcut:

\dfrac{\mathrm{d}}{\mathrm{d}x}\ln\big(g(x)\big)=\dfrac{g'(x)}{g(x)}

— “derivative of the inside over the inside”.

Final answer (a)

\dfrac{2x}{x^{2}+4}

; gradient

\dfrac{1}{2}

(b)

\dfrac{3}{\cos^{2}(3x)}

(c)

\cos x\,\mathrm{e}^{\sin x}

⚠️ Common mistake Writing

\dfrac{\mathrm{d}}{\mathrm{d}x}\ln(x^2+4)=\dfrac{1}{x^2+4}

and forgetting the

2x

🆘 Stuck on a question? Open the notes, GDC toolkit and common mistakes

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

🎯 What you must know

Standard derivatives: $x^n \to nx^{n-1}$ (any rational $n$ ), $\sin x\to\cos x$ , $\cos x\to-\sin x$ , $\tan x\to\dfrac{1}{\cos^2x}$ , $\mathrm{e}^x\to\mathrm{e}^x$ , $\ln x\to\dfrac{1}{x}$ .
Before differentiating, rewrite roots and fractions as powers: $\sqrt{x}=x^{\frac{1}{2}}$ , $\dfrac{3}{x^2}=3x^{-2}$ .
Chain rule (a function inside a function): differentiate the outside, keep the inside, then multiply by the derivative of the inside. $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{\mathrm{d}y}{\mathrm{d}u}\times\dfrac{\mathrm{d}u}{\mathrm{d}x}$ .
Product rule (two functions multiplied): $y=uv \Rightarrow y'=uv'+vu'$ .
Quotient rule (one function divided by another): $y=\dfrac{u}{v}\Rightarrow y'=\dfrac{vu'-uv'}{v^2}$ . The minus sign makes the order important.
Related rates: write an equation linking the quantities, differentiate it, then use the chain rule, e.g. $\dfrac{\mathrm{d}A}{\mathrm{d}t}=\dfrac{\mathrm{d}A}{\mathrm{d}r}\times\dfrac{\mathrm{d}r}{\mathrm{d}t}$ . Substitute the numbers only after differentiating.
Trig calculus only works in radians: set the calculator to RAD before any derivative of sin, cos or tan.

Chain rule

y=g(u),\ u=f(x)\ \Rightarrow\ \dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{\mathrm{d}y}{\mathrm{d}u}\times\dfrac{\mathrm{d}u}{\mathrm{d}x}

📘 In the formula booklet

Product rule

y=uv\ \Rightarrow\ \dfrac{\mathrm{d}y}{\mathrm{d}x}=u\dfrac{\mathrm{d}v}{\mathrm{d}x}+v\dfrac{\mathrm{d}u}{\mathrm{d}x}

📘 In the formula booklet

Quotient rule

y=\dfrac{u}{v}\ \Rightarrow\ \dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{v\dfrac{\mathrm{d}u}{\mathrm{d}x}-u\dfrac{\mathrm{d}v}{\mathrm{d}x}}{v^2}

📘 In the formula booklet

Standard derivatives

\dfrac{\mathrm{d}}{\mathrm{d}x}(\sin x)=\cos x,\ \ \dfrac{\mathrm{d}}{\mathrm{d}x}(\cos x)=-\sin x,\ \ \dfrac{\mathrm{d}}{\mathrm{d}x}(\tan x)=\dfrac{1}{\cos^2 x},\ \ \dfrac{\mathrm{d}}{\mathrm{d}x}(\mathrm{e}^x)=\mathrm{e}^x,\ \ \dfrac{\mathrm{d}}{\mathrm{d}x}(\ln x)=\dfrac{1}{x}

📘 In the formula booklet

Related rates (chain rule in time)

\dfrac{\mathrm{d}A}{\mathrm{d}t}=\dfrac{\mathrm{d}A}{\mathrm{d}r}\times\dfrac{\mathrm{d}r}{\mathrm{d}t}

🧠 Learn this one

🖩 Your GDC toolkit for this subtopic

Follow the steps on your own TI-Nspire CX II. The screens show what you should see.

Gradient at a point on a Calculator page (pastes the d/dx template)

menu💻 click menu on keypad›4: Calculus›1: Numerical Derivative at a Point…

Start on a Calculator page (ctrl doc adds one).
Press menu.
Choose 4: Calculus.
Choose 1: Numerical Derivative at a Point….
A box opens. Fill in each field (press tab to move between fields), then press enter. The results appear on the screen.

The menu path

Both ≈ 2.718, so the exact answer e is right

Worked example: Question 3 — Check with the numerical derivative

The dialog asks for the variable, the value and 1st or 2nd derivative. It pastes the d/dx template; type the function inside and press enter.

Gradient at a point on a Graphs page: choose dy/dx, click the curve, type the x-value

menu💻 click menu on keypad›6: Analyze Graph›5: dy/dx

Start on a Graphs page (ctrl doc adds one).
Press menu.
Choose 6: Analyze Graph.
Choose 5: dy/dx.
Press click on the curve, type the x-value and press enter. The gradient at that point is shown.

The menu path

At x =

\dfrac{\pi}{6}

the gradient is 3

Worked example: Question 5 — Check: dy/dx on a Graphs page (RAD mode)

Find a maximum (e.g. where a rate is largest or where f′ = 0)

menu💻 click menu on keypad›6: Analyze Graph›3: Maximum

Start on a Graphs page (ctrl doc adds one).
Press menu.
Choose 6: Analyze Graph.
Choose 3: Maximum.
Move the pointer to the LEFT of the highest point and press click, then to the RIGHT and press click. The maximum point and its coordinates appear.

The menu path

Maximum at (2.718…, 0.3679…) = (e,

\dfrac{1}{e}

)

Worked example: Question 4 — Check on a Graphs page

Graph f1(x) =

\dfrac{\ln (x)}{x}

, choose Maximum, then click to the left and to the right of the top of the curve.

Set radians for trig: doc → Settings → Document Settings → Angle: Radian

Press doc.
Choose Settings, then Document Settings.
Press tab until Angle is highlighted, press click and choose Radian.
Press tab to OK and press enter. The top of the screen now shows RAD.

Set Angle to Radian

⚠️ Where students lose marks

Forgetting to multiply by the derivative of the inside: $\dfrac{\mathrm{d}}{\mathrm{d}x}\sin(5x)=5\cos(5x)$ , not $\cos(5x)$ .
Getting the quotient rule upside down: it is $vu'-uv'$ on top (bottom × derivative of top first), and $v^2$ underneath.
Substituting the numbers into a related-rates equation before differentiating — the variable becomes a constant and its derivative wrongly becomes 0.
Calculator in DEG mode for a trig derivative — every answer comes out wrong.
Losing the sign in related rates: a length that is getting shorter has a negative rate of change.

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL HL · AI 5.9 — Further differentiation: chain, product, quotient rules and related rates

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL

5.9 Further differentiation: chain, product, quotient rules and related rates

✏️ Practice questions

🎯 What you must know

🖩 Your GDC toolkit for this subtopic

⚠️ Where students lose marks