c6-8 mastering differentiation - m1 maths
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M1Maths.com C6-8 Mastering Differentiation Page 1
M1 Maths
C6-8 Mastering Differentiation
differentiating all functions fluently
Summary Lead-In Learn Solve Revise Answers
Summary
By the time you finish this module, you should be a master of differentiation, able to
differentiate any compound function, however complex.
Learn
Introduction
By the time you finish this module, you should be able to differentiate any function
you are likely to meet, however complex, and do it fluently. This is important if you
are to be successful with the next modules on differential equations and integration.
These involve anti-differentiation, which is just the reverse process of differentiation.
It will be easy if you have thoroughly mastered differentiation; you will have trouble if
you havenβt.
The Chain Rule β Function of a Function
You probably learnt to use the chain rule using u and ππ’
ππ₯ etc. We can now use this
approach on functions with exponential and trigonometric components, like this:
Letβs differentiate y = (ex + cos x)4
y = (ex + cos x)4
y = u4 u = ex + cos x
du
dy = 4u3
dx
du = ex β sin x
dx
dy =
du
dy Γ
dx
du
= 4u3 Γ (ex β sin x)
= 4(ex + cos x)3 Γ (ex β sin x)
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Then you would have gone on to using the chain rule without writing this working,
instead thinking about the something. In this case
y = (ex + cos x)4
y = something to the power of 4
yβ = the derivative of y with respect to the something multiplied by the derivative
of the something with respect to x.
= 4(ex + cos x)3 Γ (ex β sin x)
Practice
Q1 Use the chain rule to differentiate the following. Donβt forget to get the
functions into differentiable form first by expressing fractions as negative
powers and roots as fractional powers.
(a) y = sin x3 (b) y = sin (3x2 + 5x) (c) h = (sin t + 1)5
(d) y = (3x2 + 5x)8 (e) h = 43 )2(
1
xx (f) h = 21 x
(g) y = (ln x)3 (h) h = cos (2x) (i) r = ln (t β t2)
(j) y = 2cos (3t2) (k) P = esin x (l) s = 24 xe
(m) x
ysin
1 (n) t = cos (ex) (o) A =
3
5 x
Q2 Use the chain rule to differentiate the following.
(a) y = (2x2 + x)6 (b) y = (sin x)
3 (c) h = (cos r + r2)
5
(d) y = cos (4x2 + 2x) (e) h = 3)(sin
1
x (f) h = 24 t
(g) y = sin (ln x) (h) h = cos (5x3) (i) r = (t β t2)5
(j) y = 7 sin (t4) (k) P = 2sin x
(l) s = xe 2
(m) x
ycos
4 (n) t = sin (ex) (o) A =
3
3x
(p) x = ln (sin t) (q) p = xx cossin (r) y = xln
1
(s) f = 3 sin (1 β x2) (t) t = sin s2 β (cos s)2 (u) w = 3)(cos
5
t
Some of the derivatives can be simplified and it is good style to simplify them where
possible.
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For example, Q9(s) gives 3 cos (1 β x2) Γ β2x. This can be simplified to β6x cos (1 β x2).
You should simplify answers where possible from here on and answers will be given in
simplified form.
Practice
Q3 Differentiate the following, simplifying your answer where possible. Hint:
Rewrite sin2 x as (sin x)2 etc. before differentiating, but put it back into the
sin2 x form when simplifying.
(a) y = sin x4 (b) y = sin (2x2 + 1) (c) y = (cos t + t)3
(d) y = ln 3x (e) h = cos (x + 3) (f) r = ln (a + sin a)
(g) y = β4sin (3t2) (h) P = (7x2 β 2x)11 (i) s = 632 xe
(j) x
ycos
7 (k) t = 4x + 2 sin (ex) (l) A = 3
3
2 x
(m) x = ln (t2 + cos t) (n) p = 3 2nn (o) y = 24
1
x
(p) f = sin (4x β x2) (q) t = cos s2 β cos2 s (r) b = 7)(cos
2
t
The Chain Rule β Function of a Function of a Function
Suppose we need to differentiate sin π2π₯.
2x is a function; π2π₯ is a function of a function; and sin π2π₯ is a function of a function of
a function.
Differentiating a function of a function of a function is really no harder than
differentiating a function of a function.
Again we look at the outside function, in this case sine of something.
The derivative is cos of something, in this case cos π2π₯.
Then, because the something isnβt just x, we have to multiply by the derivative of the
something.
Now, in this case, the something is a function of a function and so requires the chain
rule. Its derivative is π2π₯ Γ 2.
So the whole derivative is cos π2π₯ Γ π2π₯ Γ 2, which should be simplified to 2π2π₯ cos π2π₯.
A function of a function of a function of a function can be differentiated in the same
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way. Just start with the outside function. Then the derivative of the something will be
the derivative of a function of a function of a function.
You might be able to see why this rule is called the chain rule.
Practice
Q4 Differentiate the following. Donβt forget to simplify where appropriate.
(a) y = cos π4π₯ (b) y = sin ππ₯5 (c) y = ln (cos t)2
(d) y = sin (ln 6x) (e) h = πcos2π₯ (f) r = ln (sin a2)
(g) w = β12sin (ln x)2 (h) P = (cos 4x β 2x)11 (i) h = 3π4cos3π₯ β ππ ππ2π₯
(j) r = 2 sin (πβln π₯) (k) y = ln (cos (2
βsin (π₯+1)))
You should get to the point of being able to use the chain rule fairly automatically
without having to puzzle over it or to remind yourself of what to do. Keep practising
until you get to that point.
The Product Rule
In this section you will use the product rule in conjunction with sin x, cos x, ax ex and
ln x. Remember that the product rule is used to differentiate the product of two
factors. The derivative is the product of the factors with one of them differentiated
plus the product of the factors with the other differentiated.
So to differentiate y = 4x sin x, we write
y = 4x sin x
yβ = 4 sin x + 4x cos x
Practice
Q5 Use the product rule to differentiate the following, laying out the working
as shown above.
(a) x = x3 sin x (b) y = ex sin x (c) y = (x2 + 5x) cos x
(d) y = x ln x (e) h = sin x cos x (f) r = 2x ln x
(g) y = β4et sin t (h) P = (3x2 β 5x)(4 β x) (i) s = 6ex (x2 β 2)
(j) k = x ln x (k) t = β4x cos x (l) A = 3x 2x
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Q6 Use the product rule to differentiate the following without writing any
working.
(a) y = x2 sin x (b) v = (5x3 + 1) ex (c) P = (cos t + t) sin t
(d) y = ex sin x (e) h = (x2 β 5x) ln x (f) r = (cos w + 2 sin w) sin w
(g) y = 1.05x Γ x2 (h) P = (7x2 β 2x) cos x (i) s = 4 (x3 β x) ex
(j) f = 2x2 ex (k) t = 4x sin x (l) A = + sin x
(m) x = t ln t β et (n) p = 3 2n cos n (o) y = 5 sin x cos x
(p) f = 3 ex sin x (q) t = cos s β s cos s (r) b = (3r2 β 4r)(cos r β sin r)
If you meet an expression which is a product of 3 functions, the product rule can be
adapted to yβ = uβvw + uvβw + uvwβ. This time we write out uvw 3 times, first time with
the u differentiated, second time with the v differentiated and third time with the w
differentiated.
You can probably extrapolate to the formula for a product of 4 functions, though you
might run out of letters.
Practice
Q7 Use the product rule to differentiate the following.
(a) y = x2 ex sin x (b) y = 2
x (x3 + 2x2) cos x (c) y = (x2 + 5x) e
x ln x
The Quotient Rule
The quotient rule can be used for any function which can be written as one expression
divided by another expression.
Consider y = xx
e x
22
We let the top expression (the numerator) be u and the bottom expression (the
denominator) be v. Again, we let their derivatives be uβ and vβ.
We write:
y = xx
e x
22
u = ex v = x2 + 2x
uβ = ex vβ = 2x + 2
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yβ = 2
''
v
uvvu
= 22
2
)2(
)22()2(
xx
xexxe xx
= 22
2
)2(
)2(
xx
ex x
As with the product and chain rules, you might be comfortable doing this without the
working, though as the quotient rule formula is more complex and less symmetrical, it
can be worth continuing to use it.
Practice
Q8 Use the quotient rule to differentiate the following.
(a) y = π₯2
π₯+3 (b) y =
π₯2β π₯
π₯2+ π₯ (c) y =
sin π₯
π₯
(d) y = ππ₯
π₯2 (e) y = 2π₯
cos π₯ (f) y =
ln π₯
π₯+1
(g) y = sin π₯
cos π₯ (h) y =
cos π₯
sin π₯ (i) y = tan x
(j) y = 4π₯
ln π₯ (k) y =
ln 2π₯
ln π₯ (l) y =
ππ₯
ln π₯
(m) y = 4π₯2β 6π₯
π₯2+3 (n) y =
π₯2
π₯+3 (o) y =
2 ln π₯
1+sin π₯
Mixed Practice
The questions in the next exercise require the chain rule, product rule or quotient
rule. They will give you practice in choosing as well as more practice in using the
rules.
Practice
Q9 Use the chain rule, product rule or quotient rule to differentiate the
following.
(a) y = sin x4 (b) y = x2 sin x (c) s = 632 xe
(d) y = ln π₯
π₯3 (e) h = ex cos x (f) r = ln (a + sin a)
(g) y = β4 sin (3t2) (h) P = x9 + (7x2 β 2x)11 (i) y = 1
(π‘+ sin π‘)
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(j) x
xy
cos
7 (k) t = 4x + 2 sin (ex) (l) A = 8c sin c
(m) x = ln (t2 + cos t) (n) p = 3 2nn (o) y = 24
1
x
(p) f = xex (q) t = sin s2 + sin2 s (r) b = t
et
cos
2
Q10 Use the chain rule, product rule or quotient rule to differentiate the
following.
(a) y = cos (x3 + ex) (b) z = sin 6x (c) m = 4x ln x
(d) y = sin π₯
π₯ (e) h = β2ex (f) r = 4e5s
(g) y = β4t6 sin t (h) P = π₯4
3π₯5βπ₯2 (i) y = 1
(π β ln π)
(j) x
xy
cos
7 2
(k) t = 4e3x + 2sin 3x (l) A = ln c sin c
(m) x = π3π‘βsin π‘ (n) p = 4 sin r (o) y = xx 23
52
(p) f = x2ex (q) a = cos x2 + 3 cos2 x (r) x = t
t
sin
2
Rules in Combination
Some functions require more than one application of the chain, product and quotient
rules.
y = x2 sin 3x is an example. y = (ln (x2 + 2x))3 is another.
To differentiate these, we look at the overall structure: if a vertical line can separate it
into two functions multiplied together, then start with the product rule; if it is a
function over another function, then start with the quotient rule; otherwise, if it is a
function of a function, start with the chain rule
Example 1
Taking y = x2 sin 3x, this is a product, so we think uβv + uvβ
and write 2x sin 3x + x2 Γ . . . .
But, of course, differentiating sin 3x requires the chain rule, giving us cos 3x Γ 3.
So we get 2x sin 3x + x2 Γ 3 cos 3x, which can be simplified to 2x sin 3x + 3x2 cos 3x.
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Example 2
Taking y = (sin (x2 +2x))3, this is a function of a function, so we use the chain rule to
get
3(sin (x2 +2x))2 Γ the derivative of sin (x2 +2x).
But differentiating sin (x2 +2x) requires us to use the chain rule again
to get cos (x2 +2x) Γ (2x + 2).
So we get 3(sin (x2 +2x))2 Γ cos (x2 +2x) Γ (2x + 2),
which can be simplified to (6x + 6) sin2 (x2 +2x) cos (x2 +2x).
Basically, you just follow your nose and it should come out right. The following
questions will give you the practice you need.
Practice
Q11 Use the chain rule, product rule and/or quotient rule to differentiate the
following.
(a) y = x sin 2x (b) y = ln (x2 sin x) (c) s = 6x2 32 xe
(d) y = ln π₯2
π₯3 (e) h = e1+x sin 2x (f) r = π3π₯
sin 3π₯
(g) y = sin (3t2 et) (h) P = (sin 2x β 2x)5 (i) y = π2π‘
(π‘ β cos π‘)
(j) 2sin x
xy (k) t = 4x ex sin x (l) A = 3b sin (6b2 + 2b)
(m) x = ln (t4 sin t2) (n) p = 2sin nn (o) y = βππ₯
cos π₯
(p) f = 2x + 3 β x4 e6x (q) t = [ln (sin 2s)]3 (r) b = 2
5
cos
2
t
e t
Q12 Differentiate the following.
(a) p = x3 sin 5x (b) b = cos π₯
2π₯ (c) y = (cos t + 3t)3
(d) y = x ln 3x (e) h = 2x2 cos (x2 + 3) (f) r = ln (a + sin a2)
(g) y = β4t2 sin (3t2) (h) P = cos2 (7x2 β 2x) (i) s = ln (x cos x)
(j) x
xy
cos
2
(k) h = sin x + tan x (l) A = 3x 3
2 x
(m) z = tt
t
24 3
2
(n) g = xx cos (o) x = 4 ln (t2 cos t)
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(p) p = n 3 2nn (q) y = 24
2sin
x
x
(r) t = cos s2 sin2 s
(s) c = 5)(cos
2
b
b
(t) f = cos x sin (4x β x2) (u) y = cos tan
And a few word problems:
Practice
Q13 The tide height in metres at Wilhelm Bay last Saturday was given by
h = 1.8 cos Β½(t β 3) + 2.2, where t is the time in hours since midnight. How
fast was the tide rising or falling at 10 am?
Q14 The population of Nobbsville is given by p = 2212e0.0035t, where t is the
number of years since 1990. How fast was the population growing in 2017?
Q15 The concentration, c, in mg/L of menthoxinol in the blood t hours after
taking a dose is given by c = 3teβt.
(a) Find the rate at which the concentration is changing 4 hours after
taking a dose.
(b) Find the time at which the concentration is at a maximum.
(c) Find the maximum concentration.
Q16 Julieβs height in centimetres between the ages of 4 and 12 was given by
h = 65 Γ 1.1a, where a is her age in years. Her mass in kilograms is given by
m = 0.004a Γ h2. At what rate was her mass increasing on her 9th birthday.
Q17 The average price of houses in Blurbledale in thousands of dollars between
1994 and 2008 was given by p = π‘2β42π‘+1320
5+ π‘, where t is the number of years
since 1994. How fast were prices rising or falling in 2002?
Solve
Q51 Differentiate ln x Γ sin (x2 eβcos 3x).
Q52 Differentiating y = xx is a bit of a challenge. It can be done with what you know
about differentiation and some manipulation using what you know about logs.
Try it.
Q53 A pendulum swings such that its displacement, x, to the right of its equilibrium
position is given by x = 8 eβ0.04t cos 2t, where x is in metres and t is the time in
seconds. The exponential factor is there because the amplitude of the swings
decreases exponentially with time. Find an expression for the horizontal
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velocity and hence find the velocity the fourth time it is at the equilibrium
position.
Revise
Revision Set 1
Q61 The number of Daleks that have landed in the English countryside is increasing
by 20% per day. If there were 350 at 4 pm on April 5. At what rate were they
arriving at 4 am on April 16?
Q62 Differentiate:
(a) y = (x2 + 7x)8 (b) y = x2 e
x (c) y =
2π₯
π₯2+5
(d) p = r2 ln r (e) m = sin π₯
cos π₯ (f) y = sin (x2 + 2e
x)
(g) y = cos (x3 ex) + x (h) y = (x3 + 7x) sin2 3x
Q63 Arthur walks 3 km east from camp, then 4 km north. He is then 5 km from
camp. After that, he continues to walk north at 1 km/h. At what rate is his
distance from camp increasing after he has walked a further 2 km, i.e. when he
is 7 km further north than camp.
Revision Set 2
Q71 A Ferris wheel is turning such that the height of Car No 1 is given by
h = 8 sin t + 3, where h is the height in metres and t the time since it started in
minutes. What was the vertical component of Car No 1βs velocity 7 minutes
after starting?
Q72 Differentiate:
(a) y = x2 cos x (b) y = tan (c) h = 4e1β0.1t + 3t
(d) s = βπ₯
sin π₯ (e) y = sin (3 ln x) (f) y = 2 cos x ln x
(g) y = ln (x2 sin 5x) (h) y = sin(π₯2β3π₯)
24π₯
Q73 A weight hanging on the end of a string is swinging, but the amplitude of the
swings is slowly decreasing. Itβs sideways displacement t seconds after starting
is given by x = eβ0.02t cos 2t. What is the horizontal component of its velocity
when t = 12?
Revision Set 3
Q81 At what value of x is the graph of y = ln x β x horizontal?
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Q82 Differentiate:
(a) y = 7 Γ 10x Γ cos x (b) A =
ππ₯
π₯2 + sin x (c) y = 2βsin π₯3
(d) y = 24
4 7
x
x (e) h = 4 ln cos t (f) k = e
x cos x
(g) y = 3 sin x cos2 (3x + 4) (h) b = ln
t
t
cos
2
Q83 If t is the time in days since the start of April, the probability that it will have
rained by time t is given by p = ΒΌln (t + 1) Γ· cos 0.005t. At what rate is the
probability increasing when t = 20?
Answers
Q1 (a) sβ = 3π₯2 cos π₯3 (b) yβ = (6π₯ + 5)cos (3π₯2 + 5π₯) (c) yβ = 5 cos π‘(sin π‘ + 1)4
(d) yβ = 8(3x2 + 5x)7 (6x+5) (e) hβ = β4(3π₯2β2)
(π₯3β2π₯)5 (f) hβ =
π₯
β1βπ₯2
(g) yβ = 3(ln π₯)2
π₯ (h) hβ = β2sin 2x (i) rβ =
1βπ‘
π‘βπ‘2
(j) yβ = β12 sin (3t2) (k) Pβ = cosx esin x (l) sβ = 8x24 xe
(m) yβ = β cos π₯
sin2π₯ (n) tβ = βex sin (ex) (o) Aβ =
β3π₯2 ln 5
5π₯3
Q2 (a) yβ = 6(2x2 + x)5(4x + 1) (b) yβ = 3(sin x)
2 cos x (c) hβ = 5(cos r + r2)
4(2r β sin r)
(d) yβ = β(8x + 2) sin (4x2 + 2x) (e) hβ = 4)(sin
cos3
x
x (f) hβ =
24 t
t
(g) yβ = cos(ln π₯)
π₯ (h) hβ = β15x2 sin (5x3) (i) rβ = 5(t β t2)
4(1 β 2t)
(j) yβ = 28 t3 cos (t4) (k) Pβ = 2sin x
ln 2 cos x (l) sβ = β2xe 2
(m) x
xy
2cos
sin4' (n) tβ = ex cos (ex) (o) Aβ = 3x2 Γ
3
3xln 3
(p) xβ = cos π‘
sin π‘ (q) pβ =
cos π₯βsin π₯
2βsin π₯βcos π₯ (r) yβ =
β1
2π₯ ln π₯3
2β
(s) fβ = 6x cos (1 β x2) (t) tβ = 2s cos s2 β 2cos s sin x (u) wβ = 25 sin π‘
(cos π‘)6
Q3 (a) yβ = 4x3 cos x4 (b) yβ = 4x cos (2x2 + 1) (c) yβ = 3(cos t + t)3(1 β sin t)
(d) yβ = 1
π₯ (e) hβ = βsin (x + 3) (f) rβ =
1+cos π
π+sin π
(g) yβ = 24t cos (3t2) (h) Pβ = (7x2 β 2x)10(154x β 22) (i) sβ = 36x32 xe
(j) x
xy
2cos
sin7' (k) tβ = 4 + 2ex cos (ex) (l) Aβ = β9x2
3
2 xln 2
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(m) xβ = 2π‘βsin π‘
π‘2+cos π‘ (n) pβ =
1+2π
3 βπ4+2π3+π23 (o) yβ = 2)24(
4
x
(p) fβ = (4 β 2x) cos (4x β x2) (q) tβ = β2s sin s2 + 2 sin s cos2 s (r) bβ = 8)(cos
sin14
t
t
Q4 (a) yβ = β4π4π₯sin π4π₯ (b) yβ = 5x4 ππ₯5cos ππ₯5
(c) yβ = β2 tan t
(d) yβ = cos (ln 6π₯)
π₯ (e) hβ = β2sin x cos x πcos2π₯ (f) rβ =
2π
tan π2
(g) wβ = β24 ln π₯ cos (ln π₯)2
π₯ (h) Pβ = β11(4sin 4x β 2)(cos 4x β 2x)10
(i) sβ = β36sin x cos2x π4πππ 3π₯ β 2sin x cos x ππ ππ2π₯
(j) rβ = πβln π₯ cos πβln π₯
π₯βln π₯ (k) yβ =
2 cos(π₯+1) tan(2βsin (π₯+1)β )
βsin3(π₯+1)
Q5 (a) xβ = 2x2 sin x + x3 cos x (b) yβ = ex (sin x + cos x)
(c) yβ = (2x + 5) cos x β (x2 + 5x) sin x
(d) yβ = ln x + 1 (e) hβ = cos2 x β sin2 x (f) rβ = 2x ln 2 ln x +
2π₯
π₯
(g) yβ = β4et (sin t + cos t) (h) Pβ = 24x β 20 β 9x2 (i) sβ = (6x2 + 12x β 12)ex
(j) kβ = ln π₯+2
2βπ₯ (k) tβ = 4x sin x β 4 cos x (l) Aβ = 2
x (3 + 3x ln 2)
Q6 (a) yβ = x2 cos x + 2x sin x (b) vβ = (5x3 + 15x2 + 1) ex
(c) Pβ = cos2 t + t cos t + sin t β sin2 t
(d) yβ = ex (sin x + cos x) (e) hβ = (2x β 5) ln x + x β 5
(f) rβ = (2 cos w β sin w) sin w + (cos w + 2 sin w) cos w
(g) yβ = 1.05x ln 1.05 Γ x2 + 1.05
x Γ 2x (h) Pβ = (14x β 2) cos x + (7x2 β 2x) sin x
(i) sβ = (4x3 + 12x2 β 4x β 4) ex
(j) f = (2x2 + 4x) ex (k) tβ = 4(sin x + x cos x) (l) Aβ = + cos x
(m) xβ = ln t +1 β et (n) pβ = 2 cos π
3 βπ3 β βπ23sin π (o) yβ = 5 (cos2x β sin2x)
(p) fβ = 3 ex sin x + 3ex cos x (q) tβ = (s β 1)sin s β cos s
(r) bβ = (6r β 4)(cos r β sin r) β (3r2 β 4r)(cos r + sin r)
Q7 (a) yβ = 2x ex sin x + x2 e
x sin x x2 e
x cos x
(b) yβ = 2x ln 2 (x3 + 2x2) cos x + 2
x (3x2 + 4x) cos x β 2
x (x3 + 2x2) sin x
(c) yβ = (x2 + 8x + 10) ex ln x
Q8 (a) yβ = π₯2+6π₯
(π₯+3)2 (b) yβ = (2π₯β1)(π₯2+ π₯)β(2π₯+1)(π₯2βπ₯)
(π₯2+ π₯)2 (c) yβ = π₯ cos π₯ β sin π₯
π₯2
(d) yβ = (π₯β2)ππ₯
π₯3 (e) yβ = 2π₯(ln 2 cos π₯ + sin π₯)
cos2 π₯ (f) yβ =
π₯+1β π₯ln π₯
π₯(π₯+1)2
(g) yβ = 1
cos2 π₯ (h) yβ =
β1
sin2 π₯ (i) yβ =
1
cos2 π₯
M1Maths.com C6-8 Mastering Differentiation Page 13
(j) yβ = 4(ln π₯β1)
(ln π₯)2 (k) yβ =
βln 2
π₯(ln π₯)2 (l) yβ =
(π₯ ln π₯β1)ππ₯
π₯(ln π₯)2
(m) yβ = (π₯2+3)(8π₯β6)β2π₯(4π₯2β 6π₯)
(π₯2+3)2 (n) yβ =
π₯2+6π₯
(π₯+3)2
(o) yβ = 2+2 sin π₯β 2π₯ ln π₯ cos π₯
π₯(1+sin π₯)2
Q9 (a) yβ = 4x3 cos x4 (b) yβ = x2 cos x +2x sin x (c) sβ = 36x232 xe
(d) yβ = 1β3ln π₯
π₯4 (e) hβ = ex (cos x β sin x) (f) rβ = 1+cos π
π+sin π
(g) yβ = β24t cos (3t2) (h) Pβ = 9x8 + 11(7x2 β 2x)10(14x β 2) (i) yβ = β1βcos π‘
(π‘+sin π‘)2
(j) x
xxxy
2cos
sin7cos7'
(k) tβ = 4 + 2 ex cos (ex) (l) Aβ = 8sin c + 8c cos c
(m) xβ = 2π‘βsin π‘
π‘2+cos π‘ (n) pβ =
1+2π
3 β(π+π2)23 (o) yβ = 2)24(
4
x
(p) fβ = (x + 1) ex (q) tβ = 2s cos s2 + 2 sin s cos s (r) bβ = t
ttet
2cos
)cos(sin2
Q10 (a) yβ = β(3x2 + ex) sin (x3 + ex) (b) zβ = 6 cos 6x c) mβ = 4 + 4 ln x
(d) yβ = π₯ cos π₯β sin π₯
π₯2 (e) hβ = β2ex (f) rβ = 20 e5s
(g) yβ = β4t6 cos t β 24t5 sin t (h) Pβ = 4(3π₯5βπ₯2)π₯3β(15π₯4β2π₯)π₯4
(3π₯5βπ₯2)2
(i) yβ = 1βπ
π(πβln π)2
(j) x
xxxy
2
2
cos
sin7cos14'
(k) tβ = 12e3x + 6sin 3x (l) Aβ = ln c cos c +
sin π
π
(m) xβ = (3 β cos π‘) π3π‘βsin π‘ (n) pβ = cos π
4 βπ ππ3π4 (o) yβ =
22 )23(
)26(5
xx
x
(p) fβ = (x2 + 2x) ex (q) aβ = β2x sin x2 β 6 cos x sin x (r) xβ = t
tt tt
2sin
cos2sin2ln2
Q11 (a) yβ = 2x cos 2x + sin 2x (b) yβ = 2 sin π₯+π₯ cos π₯
π₯ sin π₯ (c) sβ = (36x4 + 12x)
32 xe
(d) yβ = 2β3 ln π₯2
π₯4 (e) hβ = e1+x (cos 2x + 2 sin 2x) (f) rβ =
3π3π₯(sin 3π₯ βcos 3π₯)
sin2 3π₯
(g) yβ = et (3t2 + 6t) cos (3t2 et) (h) Pβ = 10(sin 2xβ2x)4(cos 2xβ1)
(i) yβ = π2π‘(2π‘β1β2 cos π‘βsin π‘
(π‘βcos π‘)2
(j) yβ = sin π₯2β 2π₯2 cos π₯2
sin2π₯2 (k) tβ = 4ex (x cos x + x sin x + sin x)
(l) Aβ = 3 sin (6b2 + 2b) + 3b(12b + 2) cos (6b2 + 2b)
(m) xβ = (2π‘5 cos π‘2+ 4π‘3 sin π‘2)
π‘4 sin π‘2 (n) pβ = 1+2π cos π2
2βπ+sin π2 (o) yβ = cos π₯+sin π₯
2cos2π₯βππ₯ cos π₯
M1Maths.com C6-8 Mastering Differentiation Page 14
(p) fβ = 2 β x3 e6x (4 β 6x) (q) tβ = 6[ln (sin 2π )]2
tan 2π (r) bβ =
22
225
cos
sin4cos10(
t
ttte t
Q12 (a) pβ = 3x2 sin 5x + 5x3 cos 5x (b) bβ = βsin π₯ β ln π₯ cos π₯
2π₯
(c) yβ = 3(cos t + 3t)2 (3 β sin t) (d) yβ = ln 3x + 1
(e) hβ = 4x cos (x2 + 3) + 4x3 sin (x2 + 3) (f) rβ = 1+2π cos π2
π+sin π2
(g) yβ = β8t sin (3t2) β 24t3 cos (3t2) (h) Pβ = (4 β 28x) cos (7x2 β 2x) sin (7x2 β 2x)
(i) sβ = cos π₯βπ₯ sin π₯
π₯ cos π₯ (j)
x
xxxxy
2
2
cos
sincos2'
(k) hβ = cos x + 1
cos2 π₯ (l) Aβ =
3
2 x(3 β 9x3 ln x)
(m) zβ = 23
24
)24(
24
tt
tt
(n) gβ =
cos π₯βπ₯ sin π₯
2βπ₯ cos π₯ (o) xβ = 8 cos π‘β4π‘ sin π‘
π‘ cos π‘
(p) pβ = 4π3+5π4
3 β(π4+π5)23 (q) yβ = 2)24(
2sin42cos)48(
x
xxx
(r) tβ = 2 sin s cos s cos2 s β 2s sin s2 sin2 s (s) cβ = 6)(cos
cos2sin10
b
bbb
(t) fβ = βsin x sin (4x β x2) + (4 β 2x) cos x cos (4x β x2) (u) yβ = sin
Q13 Rising at 0.316 m/h
Q14 8.5 people per year
Q15 (a) β0.165 mg/L/h (b) t = 1 (c) 1.104 mg/L
Q16 175 kg/year (Julie was a whale.)
Q17 Falling $8200/year
Q51 yβ = 1
π₯ sin (x2 eβcos 3x) + ln x (cos (x2 eβcos 3x))(2x eβcos 3x β 3x2 sin 3x eβcos 3x)
Q52 yβ = xx (1 + ln x)
Q53 xβ = β8 eβ0.04t (0.04 cos 2t + 2 sin 2t); 12.84 m/s
Q61 432 per day
Q62 (a) 8(x2 + 7x)7(2x + 7) (b) (2x + x2) e
x (c) yβ =
2π₯2β4π₯+5
(π₯2+5)2
(d) 2r ln r + r (e) mβ = 1
cos2 π₯ (f) 2 (x + e
x) cos (x2 + 2e
x)
(g) β(x3 + 3x2) ex sin (x3e
x) + 1 (h) (3x2 + 7) sin2 3x + (6x3 + 42x) sin 3x cos 3x
Q63 0.919 km/h
Q71 6.03 m/s upwards
Q72 (a) yβ = 2x cos x + x2 sin x (b) yβ = 1
cos2 π₯ (c) hβ = β0.4e1β0.1t + 3
(d) sβ = sin π₯ββπ₯ cos π₯
2βπ₯ sin2 π₯ (e) yβ =
3 cos (3 ln π₯)
π₯ (f) yβ =
2 cos π₯
π₯ β 2 sin x ln x
(g) yβ = 2 sin 5π₯+5π₯ cos 5π₯
π₯ sin 5π₯ (h)
(2xβ3)cos(π₯2β3π₯)β4 ln 2 Γ sin(π₯2β3π₯)
24π₯
M1Maths.com C6-8 Mastering Differentiation Page 15
Q73 1.418 m/s in the positive direction
Q81 x = 1
Q82 (a) y = 7 Γ 10x (ln 10 Γ cos x β sin x) (b) Aβ =
(π₯β2) ππ₯
π₯3 + cos x (c) yβ =
2 cos π₯
3 βπ ππ2π₯3
(d) yβ = 2
6
)214(
)1424(4
x
xx (e) hβ = β4 tan t (f) kβ = ex (cos x β sin x)
(g) yβ = 3 cos x cos (3x + 4) Γ [cos (3x + 4) + 6 sin (3x + 4)] (h) bβ = tt
ttt
cos
sincos
Q83 0.0123/day