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www.curriculumpress.co.uk Number A2P 004
Answering A2 Exam Questions on
Differentiation
This Factsheet focuses on exam questions on differentiation. b) Methods involved in applications
Before starting it, you should know how to:-
• differentiate powers of x; To find a tangent or normal:-
• differentiate trigonometric, exponential and logarithmic functions; Find x, y coordinates at point concerned
• use the the product, quotient and chain rules; dy
• differentiate implicitly; Find and substitute in to find gradient at that point
dx
• differentiate parametric equations; dy 1
• find and distinguish between different types of stationary points; Tangent grad = value of . Normal grad = − dy
• find gradients, tangents and normals. dx
dx
Substitute in to y – y = m(x – x )
1. Reminder – the basic skills 1 1
Success in this topic requires a combination of memory and skill with
methods. Although everyone will need to both memorise some facts, To find a stationary point:-
dy
and become confident with methods, the balance between the two is Find , equate to zero and solve equation
partly down to personal preference. If you do not like working things dx
out in the exam, and prefer to memorise, then you will be better off Substitute back to find both coordinates of the point
learning more standard results – whereas if you are very quick and To determine the nature of a stationary point,
2
confident with methods, but hate rote learning, you can get by with dy
memorising relatively little. either:- find dx2 , substitute x-value in. If result is
a) Methods of differentiating • positive, point is a minimum,
• negative, it is a maximum,
General methods: • zero, it could be maximum, minimum or inflection
dy du dv dy
Product rule: If y = uv, then = v + u or Use to check the gradient each side of stationary
dx dx dx dx
du dv point. If it is:-
vu− • − 0 +, it is a minimum,
The first key box below lists the standard derivatives that everyone
Quotient rule: If y = u , then dy = dx dx • + 0 −, it is a maximum
2
needs to memorise – hopefully most of them will already be in your
head! v dx v • + 0 + or – 0 − , it is a point of inflection
Chain rule: dy = dy × du c) Facts to learn
dx du dx The first key box contains the standard results everyone has to know.
(or, informally: “differentiate a bracket by differentiating Hopefully they will mostly already be in your memory! Some may be
normally and multiplying by differential of bracket” in your formula book – but you should still learn them for speed.
Standard derivatives
These methods are generally applicable – i.e. you can expect to use function derivative
them in any type of question in differentiation. There are also some n n – 1
specific methods (actually all applications of the chain rule) that you x nx
x x
have to use in just specific types of question. e e
lnx 1/
x
sinx cosx
Methods for particular question types cosx -sinx
dy
Parametric equations: dy = dt The second key box gives some trigonometric derivatives that can be
dx dx obtained from the sin and cos results using the quotient rule. Unless
dt you are exceptionally fast with the quotient rule, you should also
Inverse functions/you are given x in terms of y: dy = 1 memorise these. Memorising them will also help you with integration.
dx dx Further trigonometric derivatives
dy function derivative
Neither x nor y the subject:- differentiate function of y”normally”
tanx sec2x
then multiply by dy secx secxtanx
dx cosecx −cosecx cotx
2
cotx -cosec x
1
Answering A2 Exam Questions on Differentiation Maths Factsheet
Exam Hint:- These questions are to help you decide whether to learn the final box.
You can also be asked to prove the additional trig results Differentiate the following:-
(use tan= sin/cos, sec= 1/cos, cosec= 1/sin, cot= cos/sin). 10 tanx – secx
(sinx + cosx) e ln(cosecx + cotx)
However, unless you are told to prove them, you can use x 2
them without proof – and it’s much quicker to know them sin(e + 2x) cos(tanx) tan(lnx – x )
4 -2 sinx
than to have to work them out. sec(3x ) cosec(4 – x ) cot(e )
2. What is on each module?
Tip The table shows which aspects of differentiation are on C3 and C4 for
If you get confused when remembering the signs of trig each specification.
derivatives, remember:-
the ones that start “co..” differentiate to give a negative, Spec. C3 C4
all the others give a positive Edexcel exponentials + logs implicit
trigonometric parametric
x x
product, quotient & chain a diff. to a lna
The third key box gives some derivatives that can be obtained from dy = 1 connected rates of change
chain rule. Decide whether you would rather
those above using the dx dx
learn them, or work them out for yourself in an exam, by trying the dy
questions that follow. AQA exponentials + logs implicit
trigonometric parametric
function derivative product, quotient & chain
n n – 1
(ax+b) na(ax+b) dy
ax+ b ax+ b 1
e ae = dx
ln(ax+ b) a/ dx
(ax + b) dy
sin(ax + b) acos(ax + b)
cos(ax + b) −asin(ax + b) OCR exponentials + logs trigonometric
2(ax + b)
tan(ax + b) asec product, quotient & chain implicit
sec(ax + b) asec(ax + b)tan(ax + b) dy = 1 parametric
cosec(ax + b) -acosec(ax + b)cot(ax + b) dx dx
cot(ax + b) -acosec2(ax + b) dy
connected rates of change
Questions (answers at top of next column). Try doing these using the MEI exponentials + logs parametric
chain rule, to see whether you should learn the results above: product, quotient & chain
Differentiate the following:- dy = 1
sin2x cos3x tan(x + 2) dx dx
sec(3x – 4) cosec( ½ x) cot(2 – x) dy
(3 – 2x) ½ e ¼ x – 3 ln(2x + 3)
connected rates of change
The final key box in this section gives even more general forms of the implicit
above derivatives, where the expression in the bracket is any function WJEC exponentials + logs -
at all – written generally as f(x) – eg for sin(2x2 – 4), f(x) = 2x2 – 4. product, quotient & chain
Remember f′ (x) means the differential of f(x). dy = 1
dx dx
function derivative dy
n n – 1 connected rates of change
(f(x)) nf′ (x)(f(x))
2 12 2 11 implicit
eg (x – 4x) 12(2x – 4)(x – 4x)
f(x) f(x) nd
e f′ (x)e parametric including 2
sin2x sin2x derivatives
eg e 2cos2x e
ln( f(x)) f′ (x)/ CCEA exponentials + logs implicit
f(x)
(cosx – sinx) trigonometric parametric
eg ln(sinx + cosx) /(sinx + cosx)
nd
sin(f(x) f′ (x)cos(f(x) product, quotient & chain Both including 2
x x x dy derivatives
eg sin(e ) e cos(e ) = 1dx
cos(f(x) −f′ (x)sin(f(x) dx
3 2 3 dy
eg cos(3+ x ) -(3x)sin(3 + x)
tan(f(x) f′ (x)sec2(f(x) This information is mainly useful in telling you what will not be asked
3x 3x 2 3x on C3. In C4 you can be asked to use anything from earlier modules –
eg tan(e ) 3e sec (e )
This formula is just another way of saying that there is a common including AS ones. Although C1-C3 material won’t be the focus of an
sec(f(x) f′ (x)sec(f(x)tan(f(x)
eg sec(lnx) 1/ sec(lnx)tan(lnx) entire question on C4, it certainly could be worth quite a few marks.
difference (i.e. we always get the same when we find the difference of
x
a term and the previous one). However, it can be useful to avoid too
cosec(f(x) - f′ (x)cosec(f(x)cot(f(x) This Factsheet will focus only on material that is common to all
2 2 2
many different variables, in type 8 questions below.
eg cosec(2x – x ) -(2 – 2x)cosec(2x – x )cot(2x – x ) specifications
cot(f(x) - f′ (x)cosec2(f(x)
-2 -3 2 -2 Answers to questions
eg cot(x ) -(-2x )cosec (x )
2cos2x -3sin3x sec2(x + 2)
3sec(3x – 4)tan(3x – 4) -½ cosec(½ x)cot(½x) cosec2(2– x)
NB - ½ ¼ x − 3 2
• If you decide to learn the final box, you don’t need to -(3 – 2x) ¼ e /(2x + 3)
learn the third one as well – the last one covers it 9 2 tanx – secx
10(sinx + cosx) (cosx–sinx) (sec x–secxtanx)e -cosecx
x x 2 1 2 2
• The final box can be summarised as “differentiate as if it (e + 2)cos(e + 2x) -sec x sin(tanx) (/x–2x)sec (lnx–x )
3 4 4 -3 -2 -2 sinx 2 sinx
was just x, then times by differential of bracket” 12x sec(3x )tan(3x ) -2x cosec(4–x )cot(4–x ) -cosxe cosec (e )
2
Answering A2 Exam Questions on Differentiation Maths Factsheet
3. Approaching questions Such questions may also ask you to
The main problems students tend to have are:- • use trigonometric identities to assist with proofs
• not knowing which method of differentiation to use; • use the laws of logs
• in some cases, not knowing they have to differentiate. • use the fact that integration is the inverse process of differentiation
This section of the Factsheet aims to deal with these. to find an integral
• sketch a curve
(a) Which method? • find whether/where a tangent/normal meets the curve again
First, it makes sense to ask yourself which module you are doing! For • integrate to find an area or volume
most boards, for example, parameters are not on C3 – so you know • find the Cartesian equation (for parametrics)
that method won’t be required. The table below shows some “cues” These parts of the questions are covered in separate Factsheets
that questions give you as to which method of differentiation to use.
Question content You use… We will concentrate here on types 2 – 5, as type 1 questions are
You have something that could be Product rule covered within these, and type 6 are a matter of learning only.
written as two things multiplied
together – i.e. the x occurs in more Try doing each question before looking at the solution – and check
than one place. your work against the “common errors” listed afterwards!
You have something that looks like a Quotient rule
fraction, with x in both the top and eg 1. A curve has equation x = 2y2 – 3y.
bottom. 1
Find the point at which its gradient is /5
You have something in brackets, but Chain rule
x only occurs once The word “gradient” tells us we must differentiate.
In some questions you will have to use simple cases of the chain rule The fact x is given in terms of y tells us to use dy =1 dx
together with product or quotient rule. You’ll see this once you have dx dy
written down your “u” and “v” for the product and quotient rule – dx
one or both of u and v may need to be differentiated using chain rule. So, we find = 4y – 3
The question refers to parameters, dy dy
you have equations x = … & y = …, dy = dt dy 1
both in terms of t or θ or similar. dx dx So = (make your method clear to get follow-through)
dt dx 43y −
You have an equation where x is dy dy dx 1 11
= =1 We want the gradient to be /5. So put =
given in terms of y, or you are asked dx dx dy 54y−3
-1
to differentiate something like sin x. (for the inverse functions, you Rearranging gives 5 = 4y – 3 , so y = 2
change it to siny = x first) Putting y = 2 in the equation gives x = 2, so the point is (2, 2)
You have an equation with x and y all Implicit differentiation
mixed up together. (you may also need to use Common Errors
product rule if you have • Forgetting to do 1 dx
something like xy) dy
• Forgetting to find x as well as y
(b) How do I know to differentiate? dy
In some cases, the question tells you clearly to do so – you see words • Substituting 1/5 in as y, instead of putting it =
dx
such as “derivative” and/or symbols like f ′(x) or dy . Other than that,
dx
the following words tell you:- eg 2. A curve has equation y = x2ln(2x). Find its gradient at the point
• gradient where x = ½ e2, giving your answer in terms of e
• maximum/minimum/inflection/stationary point/turning point
• tangent/normal Again, “gradient” tells us to differentiate
• rate of change/ rate of increase/rate of decrease The fact that x occurs in two places at once means we will be using the
product rule:-
u = x2 v = ln(2x)
4. Types of question du dv 21
So = 2x
The main types are:- ==
dx dx 2x x
2
dy dy dy 1
⎛⎞
1. Just asking you to find (and maybe ) 22
So = 2xln(2xx)+=2xln(2x)+x
dx 2 ⎜⎟
dx dx x
⎝⎠
dy
2. Using to find a gradient, or the point where the gradient has a We want the gradient when x = ½ e2, so substitute it in:
dx
certain value dy 222222
=+2½ leen2½ ½ e=eln(e)+½ e
()()
dy dx ( )
3. Using to find a tangent or normal We then have to use the fact that ln and exponential are opposites, so
dx
dy ln(e2) = 2. Hence dy = 2½e2
4. Using to find a stationary point for a curve, or a practical dx
dx
problem, and determine its nature Common Errors
2
dy • Differentiating ln(2x) incorrectly as / (if you did this
5. Using to find a rate of change you maybe need to learn the rules!); x
dx • Putting everything into the calculator although it wants
6. Proofs of standard results, such as differentiating trigonometric e answer in terms of e.
functions or inverse trigonometric functions. th
3
Answering A2 Exam Questions on Differentiation Maths Factsheet
2 eg 5. Find the tangent(s) to the curve -x2+ xy + y2 = 1 at the point(s)
eg 3. Find the tangent to the curve y = sin x at the point where x=¼π
where x = 1
“Tangent” tells us to differentiate.
2 Again, we know we must differentiate because of “tangent”.
We don’t have an x in more than one place in sin x – but nor do we Since we have x and y all mixed up together, we must use implicit
seem to have a bracket. So what method to use?
2 2 differentiation.
The answer is to recall that sin x means (sinx) .
This means we need to use the chain rule to differentiate it We differentiate the entire equation term by term, remembering that
(we could write it as sinx × sinx and use the product rule, but that will
3 4 dy
take longer, and wouldn’t work for sin x, sin x etc) we differentiate a function of y “normally”, and then multiply by .
dx
Writing the chain rule out in full (not compulsory, but advisable if you
feel at all unsure): However, xy is a product. So we will work out what that differentiates
2
y = u u = sinx to first: u = x v = y
dy= 2u du = cosx du = 1 dv = 1 dy so xy differentiates to y + x dy
du dx dx dx dx dx
dy = dy × du = 2u cosx = 2sinx cosx So differentiating the equation:
dx du dx -2x + y + x dy + 2y dy = 0
dx dx
For the tangent, we need the gradient at the point, and its coordinates. To find the gradient, we need dy . We also need y to find this.
Gradient = 2sin(¼π)cos(¼π) = 1 dx
2
y = sin (¼π) = ½ So substitute x = 1 in the original equation:
So tangent is y – ½ = 1(x − ¼π) or y = x + ½ - ¼π -1 + y + y2 = 1 so y2 + y - 2 = 0.
Common Errors This gives (y + 2)(y − 1) = 0. So y is 1 or –2
2 Exam Hint:-
• Differentiating sin x as 2sinx; For the
o You should not be surprised by there being two answers
• Working in degrees, and putting in ¼ π as 45 in from the wording of the question!
equation of tangent;
• Putting in the gradient of the tangent as “2sinxcosx” We must work everything out for the two points separately:-
rather than as a number. dy dy dy
Find at (1, 1): substituting in gives –2 + 1 + + 2 = 0
x2 dx dx dx
eg 4. The curve C has equation y = 1 1 1
x+2 This gives gradient = /3, so tangent is y – 1 = /3(x – 1) or y = /3(x+2)
(a) Find the normal to the curve at the point where x = 2 Find dy at (1, -2): substituting in gives –2 − 2 + dy − 4 dy = 0
(b) Find the coordinates of the point where the normal crosses the dx dx dx
curve again. 4 4 2
This gives gradient = - /3, so tangent is y+2 =- /3(x–1) or y =- /3(2x+1)
Common Errors
(a) “Normal” tells us to differentiate dy
Because this looks like a fraction, we use the quotient rule: • Not putting the in when differentiating terms in y ;
u = x2 v = x + 2 dx
du = 2x dv = 1 • Not treating xy as a product;
dx dx • Assuming there is an error when two answers appear.
2
dy (2x +−)2x x (1)
So = 2 eg 6. A curve has parametric equations x = t2 – 1, y = t3 – t
dx +
(2x ) Find the coordinates of the turning point(s) of the curve
Exam Hint:-
dy Because we have the word “parameters” and two equations, we know
Since you are only going to substitute a value into dx , it
dy
is not worth wasting time multiplying out brackets etc. we must use dy = dt
dy dx dx
When x = 2, y = 1, dx = ¾ dt
4 2
So gradient of normal is – /3 dx dy 2 dy 31t −
4 1 dt = 2t, dt = 3t − 1 so dx = 2t
So equation of normal is y – 1 = - /3(x – 2) or y = /3( -4x + 11)
(b) We need intersection points of curve and normal – equate: 2
dy 3t −1
2 For a turning point, we want = 0, so = 0
x−+4x11 2 dx 2t
= which gives: 3x = (x + 2)(-4x + 11)
x+23 2 1
Rearranging and solving: 7x2 – 3x + 22 = 0 This gives 3t = 1 so t = ± 3
11
(7x – 11)(x – 2) = 0 so x = /7 (x = 2 was original point)
121 1 2 2 1 2 2
y = /175 t = ⇒ x = , y = − & t = −⇒ x = , y =
3 −3 33 3 −3 33
Common Errors Common Errors
• Just differentiating top and bottom, rather than using • Wasting time trying to get y in terms of x ;
quotient rule; • Not getting the ± roots;
• Not doing “-1 over…” for gradient of normal;
• Not finding y coordinate of point of intersection. • Working in terms of decimals rather than surds.
.
4
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