4
Differentiation
4.1
Derivatives of some basic functions
The following table lists the functions f(x) and their derivatives f′ (x).
f(x)
ex
ln x
sin x
cos x
tan x
tan−1 x
f′ (x)
ex
1
x
cos x
− sin x
sec2 x
1
1+x2
Proof for ex : The definition of e as lim (1 + x1 )x means that when x is large, e ≈ (1 + x1 )x .
x→∞
1
Rearranging this gives
limits, lim
h→0
eh −1
h
e x −1
1
x
≈ 1. Let h = x1 . Then when h is small,
eh −1
h
≈ 1. In terms of
= 1. Let y = ex . By the limit definition of the derivative,
dy
dx
x+h −ex
= lim e
h
= ex lim e h−1 = ex
h
h→0
h→0
Proof for ln x: Let y = ln x. Then x = ey and
dy
dx
=
1
ey
dx
dy
= ey . Since
=
dy
dx
=
1
dx
dy
, therefore
1
x
Proof for sin x: The proof uses the following results: lim
h→0
sin h
h
= 0 and lim
h→0
cos h−1
h
= 0 (proof
not given). Let y = sin x. By the limit definition of the derivative,
dy
dx
x
= lim sin(x+h)−sin
= lim sin x cos h+sinh h cos x−sin x = sin x lim cos hh−1 + cos x lim sinh h = cos x
h
h→0
h→0
h→0
Proof for tan−1 x: Let y = tan−1 x. Then x = tan y and
dy
dx
=
1
sec2 y
=
1
1+tan2 y
=
dx
dy
h→0
= sec2 y. Then
1
1+x2
Remark: f′ (x) is the gradient function of f(x). For trigonometric functions, the functions are
assumed to be applied to angles in radians. For example, the gradient of the curve y = sin x
at the point where x = 1 is equal to cos 1 = 0.5403 (not 0.9995).
Exercise 4.1: Find an expression for
i) y = e2x−1
iv) y = 2 ln(x2 − 2x − 2)
vii) y = tan2 2x
x) y = sec 2x
xiii) y = e2 cos x
dy
.
dx
1 2
ii) y = 3e− 2 x
v) y = (ln x)2
√
viii) y = cos x
xi) y = tan−1 ( 21 x)
xiv) y = ln(1 + cos 2x)
1
iii) y = ln(2x + 3)
vi) y = sin( 13 π − 3x)
ix) y = cot x
xii) y = tan−1
√
x
xv) y = tan−1 (cot x)
Exercise 4.2: A curve has equation y = 23 ln(1 + 3 cos2 x) for 0 ≤ x ≤ 21 π.
i) Express
dy
dx
in terms of tan x.
ii) Hence find the x-coordinate of the point on the curve where the gradient is −1. Give your
answer correct to 3 significant figures.
Exercise 4.3: The equation of a curve is y = 3 sin x + 4 cos3 x.
i) Find the x-coordinates of the stationary points on the curve in the interval 0 < x < π.
ii) Determine the nature of the stationary point in this interval for which x is least.
4.2
Product and quotient rule
Let u and v be functions of x. The product rule states that
d
(uv)
dx
dv
= u dx
+ v du
dx
Proof : Let y = u(x)v(v). Then by the limit definition of the derivative,
dy
dx
u(x+h)v(x+h)−u(x)v(x)
h
h→0
= lim
= lim
h→0
u(x+h)v(x+h)−u(x+h)v(x)+u(x+h)v(x)−u(x)v(x)
h
v(x+h)−v(x)
h
h→0
= lim u(x + h) lim
h→0
u(x+h)−u(x)
h
h→0
+ v(x) lim
dv
= u(x) dx
+ v(x) du
dx
Exercise 4.4: The expression f(x) is defined by f(x) = 3xe−2x . Find the exact value of f′ (− 21 ).
1
Exercise 4.5: Find the equation of the tangent to the curve y = x 2 ln x at the point where
x = 1, giving your answer in the form y = mx + c.
Exercise 4.6: Find the exact coordinates of the stationary point on the following curves.
1
i) y = (x + 1)e− 3 x
ii) y = x2 ln x
Exercise 4.7: The diagram shows the curve and its maximum point M . Find the exact
x-coordinate of M .
1 2√
i) y = x2 e2−x
ii) y = e− 2 x 1 + 2x2
2
Exercise 4.8: For the following curves, find the x-coordinate of the stationary point in the
interval 0 < x < 21 π, giving your answer correct to 3 significant figures.
ii) y = sin2 2x cos x
i) y = sin x cos 2x
iii) y = 5 sin3 x cos2 x
iv) y = e2 sin x cos x
1
Exercise 4.9: The diagram shows the curve y = 10e− 2 x sin 4x for x ≥ 0. The stationary
points are labelled T1 , T2 , T3 , ... as shown.
i) Find the x-coordinates of T1 and T2 , giving each x-coordinate correct to 3 decimal places.
ii) It is given that the x-coordinate of Tn is greater than 25. Find the least possible value of
n.
Exercise 4.10: The equation of a curve is y = e−2x tan x, for 0 ≤ x < 12 π.
dy
i) Obtain an expression for dx
and show that it can be written in the form e−2x (a + b tan x)2 ,
where a and b are constants.
ii) Explain why the gradient of the curve is never negative.
iii) Find the value of x for which the gradient is least.
Let u and v be functions of x. The quotient rule states that
d u
dx v
=
dv
v du
− u dx
dx
v2
Remark: The proof is left as an exercise.
Exercise 4.11: Find the gradient at the point where the curve crosses the y-axis for y =
Exercise 4.12: Find the exact coordinates of the point on the curve y =
gradient of the tangent is equal to 41 .
Exercise 4.13: The equation of a curve is y =
negative.
1+e−x
,
1−e−x
x
1+ln x
for x > 0. Show that
1+x2
.
1+e2x
at which the
dy
dx
is always
sin x
Exercise 4.14: The equation of a curve is y = 1+cos
, for −π < x < π. Show that the gradient
x
of the curve is positive for all x in the given interval.
3
Exercise 4.15: Find the exact coordinates of the stationary points on the following curves.
i) y =
iii) y =
x2
1+x3
ii) y =
(ln x)2
x
iv) y =
2−sin x
cos x
for x > 0
e2x
x3
for − 12 π < x < 12 π
Determine whether the point is a
maximum or a minimum point
Determine whether the point is a
maximum or a minimum point
Exercise 4.16: The diagram shows the curve y =
q
1−x
.
1+x
dy
, obtain an expression for dx
in terms of x. Hence show that
i) By first differentiating 1−x
1+x
√
the gradient of the normal to the curve at the point (x, y), is (1 + x) 1 − x2 .
ii) The gradient of the normal to the curve has its maximum value at the point P shown in
the diagram. Find, by differentiation, the x-coordinate of P .
4.3
Parametric differentiation
If x and y are functions of a parameter t, then the pair of functions is called a parametric
equation.
For each value of t, there is an x-value and y-value, which are the x-and y-coordinates of a
point. The graph consists of all points (x, y) as t varies. For example:
ˆ x = cos t, y = sin t,
ˆ x = t2 − 1, y = t,
ˆ x = sin 2t, y = cos 3t,
for 0 ≤ t ≤ π
for t ∈ R
for 0 ≤ t ≤ 2π
y
y
y
t= 1
π
2
t=0
t=1
7π
t= 1
π, 6
6
x
t=0
x
t=π
t=0
x
t = −1.5
The gradient function of the curve is
rule,
dy
,
dx
which is a function in the parameter t. By the chain
dy
dx
=
dy
dt
.
dx
dt
The gradient of a parametric curve at a point is equal to
corresponding to the point.
4
dy
dx
evaluated at the value of t
Exercise 4.17: Find the gradient of the curve with the following parametric equations.
i) x =
t
,
2t+3
y = e−2t ,
ii) x =
at the point for which t = 0
iii) x = ln(2t + 3), y =
4t
,y
2t+3
= 2 ln(2t + 3),
at the point for which x = 1
iv) x = sin t + cos t, y = sin3 t + cos3 t,
3t+2
,
2t+3
at the point where it crosses the y-axis
at the origin
Exercise 4.18: The parametric equations of a curve are x = 2t + sin 2x, y = 1 − 2 cos 2t, for
− 12 π < t < 21 π.
i) Show that
dy
dx
= 2 tan t.
ii) Hence find the x-coordinate of the point on the curve at which the gradient of the normal
is 2. Give your answer correct to 3 significant figures.
Exercise 4.19: The parametric equations of a curve are x = t + cos t, y = ln(1 + sin t), where
− 21 π < t < 21 π.
i) Show that
dy
dx
= sec t.
ii) Hence find the x-coordinates of the points on the curve at which the gradient is equal to
3. Give your answers correct to 3 significant figures.
Exercise 4.20: The parametric equations of a curve are x = 2 sin θ+sin 2θ, y = 2 cos θ+cos 2θ,
where 0 < θ < π.
i) Obtain an expression for
dy
dx
in terms of θ.
ii) Hence find the exact coordinates of the point on the curve at which the tangent is parallel
to the y-axis.
Exercise 4.21: The parametric equations of a curve are x = ln(tan t), y = sin2 t, where
0 < t < 12 π.
i) Express
dy
dx
in terms of t.
ii) Find the equation of the tangent to the curve at the point where x = 0.
Exercise 4.22: The diagram shows the curve with parametric equations x = tan θ, y = cos2 θ,
for − 21 π < θ < 12 π.
i) Show that the gradient of the curve at the point with parameter θ is −2 sin θ cos3 θ.
ii) The gradient of the curve has its maximum value at the point P . Find the exact value of
the x-coordinate of P .
5
4.4
Implicit differentiation
An equation (in x and y) which is not in the form y = f(x) is called implicit, e.g. y 2 = x.
√
√
Remark: The equation y 2 = x can be written as y = ± x. However, ± x is not a function.
The graph of y against x consists of all points with coordinates (x, y) satisfying the implicit
equation. For example:
ˆ y 2 = x3 − x
ˆ y 2 = x3 − x + 1
dy
,
dx
The gradient function of the curve is
ˆ (x2 + y 2 − 1)3 = x2 y 3
which is a function in both x and y.
Given an implicit equation (in x and y), to find
with respect to x.
dy
,
dx
differentiate both sides of the equation
When differentiating an expression in terms of y with respect to x, differentiate it with respect
dy
. This is a consequence of the chain rule.
to y then multiply the derivate with the term dx
d
f(y)
dx
Then rearrange the equation to find
=
d
f(y)
dy
×
dy
dx
dy
.
dx
For example, consider the implicit equation y 2 = x3 − x + 1. Differentiating both sides of the
equation with respect to x gives
d
(y 2 )
dx
d
(y 2 )
dy
×
dy
dx
=
d
(x3
dx
− x + 1)
= 3x2 − 1
dy
2y dx
= 3x2 − 1
dy
dx
=
3x2 −1
2y
If the implicit equation contains a product involving both x and y terms, use the product rule
where u is the x term and v is the y term.
For example,
d
(xy 2 )
dx
dy
= x(2y dx
) + y2.
The gradient of an implicit curve at a point (a, b) is equal to
6
dy
dx
evaluated at x = a and y = b.
Exercise 4.23: Find the gradient of the following curves.
i) x3 + 3xy 2 − y 3 = 1 at (1, 3)
ii) 3x2 − 4xy + y 2 = 45 at (2, −3)
iii) 3e2x y + ex y 3 = 14 at (0, 2)
iv) 2x3 + 5xy + y 3 = 8 at the y-axis
Exercise 4.24: The equation of a curve is x ln y = 2x + 1.
i) Show that
dy
dx
= − xy2 .
ii) Find the equation of the tangent to the curve at the point where y = 1, giving your answer
in the form ax + by + c = 0.
Exercise 4.25: The variables x and y satisfy the relation sin y = tan x, where − 21 π < y < 12 π.
dy
Show that dx
= cos x√1cos 2x .
Remark: To find coordinates of stationary points, obtain an equation from
simultaneously with the equation of the curve, e.g. by substitution.
dy
dx
= 0 and solve
Exercise 4.26: The diagram shows the curve with equation x3 + xy 2 + ay 2 − 3ax2 = 0, where
a is a positive constant.
The maximum point on the curve is M . Find the x-coordinate of M in terms of a.
dy
Exercise 4.27: For the following curves, show that dx
is the expression given. Then find the
coordinates of the points on the curve where the tangent is parallel to the x-axis.
i) 2x4 + xy 3 + y 4 = 10
dy
dx
3
ii) x3 − 3x2 y + y 3 = 3
3
8x +y
= − 3xy
2 +4y 3
dy
dx
=
x2 −2xy
x2 −y 2
Exercise 4.28: The equation of a curve is xy(x − 6y) = 9a3 , where a is a non-zero constant.
Show that there is only one point on the curve at which the tangent is parallel to the x-axis,
and find the coordinates of this point.
Exercise 4.29: The equation of a curve is 3x2 − 4xy + y 2 = 45. Show that there are no points
on the curve at which the gradient is 1.
dy
Exercise 4.30: For the following curves, show that dx
is the expression given. Then find the
coordinates of the points on the curve where the tangent is parallel to the y-axis.
i) x3 + 3xy 2 − y 3 = 5
dy
dx
=
ii) ln(xy) − y 3 = 1
dy
dx
x2 +y 2
y 2 −2xy
7
=
y
x(3y 3 −1)