Differentiation
February 22, 2017
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Introduction
Differentiation of a function, is sometime called the derivative of the functions. In
this course, we assumed that the functions that we used are differentiable, meaning that the derivative functions exist. Particularly, in this chapter, we will learn
how to differentiate hyperbolic functions, inverse trigonometric functions, and inverse
hyperbolic functions.
To review, let us look at the techniques of differentiations that you should have
mastered by now.
Differentiation of Sums
Let u and v be differentiable functions of x. If y = u + v, then
du dv
dy
=
+ .
dx
dx dx
Example 1: Differentiate the following functions with respect to x.
a) y = 4x2 + 5x3 .
b) y = 6x7 − 3x−8 .
c) y = (2x2 − x)2 .
d) y = x2 + cos x.
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Differentiation of Products
Let u and v be differentiable functions of x. If y = uv, then
dy
dv
du
=u +v .
dx
dx
dx
We usually refer this as ’product rule’.
Example 2: Differentiate the following functions with respect to x.
a) y = x3 cos x.
b) y = 5x sin x.
c) y = (2 − x3 )(x3 + x − 1).
d) y = (x + 1)(x2 − x + 1).
Differentiation of Quotients
Let u and v 6= 0 be differentiable functions of x. If y =
u
, then
v
dv
du
−u
v
dy
= dx 2 dx .
dx
v
We usually refer this as ’quotient rule’.
Example 3: Differentiate the following functions with respect to x.
1
.
1 − sin x
cos x
b) y =
.
1 − cos x
a) y =
2
c) y =
1
.
3x − 5
d) y =
2x − 1
.
x+1
The Chain Rule
In doing differentiation, often times we will need to find the derivative of a composite
function and/or complicated functions. The chain rule is a very useful technique in
dealing with those functions. It is emphasised that in this course, we will have to use
chain rule in most of the time when we try to find the derivative of the hyperbolic
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functions. We will also make use of chain rule again when we use the formula to
find the derivative of the inverse trigonometric, and inverse hyperbolic functions.
Differentiation of parametric equation will also use chain rule. The chain rule is
defined as below.
If g is differentiable at x and f is differentiable at the point g(x), then the composite function f · g is differentiable at x. In other words, if y = f (g(x)) and u = g(x)
then
dy du
dy
=
× .
dx
du dx
Example 4: Differentiate the following functions with respect to x.
a) y = (x3 + 6x − 3)20 .
b) y =
√
3 − 2x.
c) y = sin(2x − 5).
d) y = cos3 (x3 − 5x2 + 2).
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Differentiation of Implicit Functions
In all the previous example, we have done the differentiation explicitly, such that the
function y is already given and expressed in term of x only. There are times when
the given function is in terms of F (x, y) = 0, and to express y in term of x is a
tedious work. The best way to deal with such function is to differentiate it implicitly.
Example 5
1. Differentiate the following functions implicitly with respect to x.
a) y + xy = 3.
b) 3x2 − xy + 3y = 7.
c) 3y 2 − 2x2 = 2xy.
d) x3 + y 3 = a3 .
e) x2 + y 2 = 1.
dy
d2 y
+
2
+ 10y = 0.
dx2
dx
2
d2 y
dy
-1
= 0.
3. If y = tan (sinh x), show that 2 + tanh x
dx
dx
2. If y = e−x cos 3x, show that
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4
Differentiation of Logarithmic Functions
As you must have been familiar by now, if we have y = ln u, where u = u(x)(function
of x), the derivative is given as
dy
dy du
1 du
=
×
= × .
dx
du dx
u dx
dy
1
For example if we have y = ln x, then
= . Logarithmic differentiation is also
dx
x
useful if we are dealing with complicated function that required both product rule
and quotient rule. Logarithmic differentiation, combined with implicit differentiation
can deal with the messy product elegantly. Additionally, logarithmic differentiation
is also useful when we are trying to find the derivation of an exponential function.
Example 6: Differentiate the following functions with respect to x.
a) y = xx .
b) y =
(x + 1)2 (x + 2)
.
(x + 3)3
c) y = ln(x3 + 1).
d) y = ln(cosec x).
e) y = log10 x2 .
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Differentiation of Parametric Functions
In some cases, implicit function can be expressed in term of parameters. For example
x = t2 and y = t3 + t,
is a parametric equations of a curve with parameter t which is equivalent to the
implicit functions y 2 = x(x + 1)2 . The differentiation of parametric functions can be
done by using chain rule, where if y = y(t), x = x(t), then
dy
dy
dt
dt
1
=
× , where
=
.
dx
dt
dx
dx
dx/dt
The results are the same if we first write y in term of x only and to the differentiation
afterwards.
Example 7: Differentiate the following functions with respect to x.
dy
a) Find
if x = 2t and y = 4 − 4t − 4t2 .
dx
dy
d2 y
t
b) Find
and dx
2 if x = e and y = sin t.
dx
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Differentiation of Hyperbolic Functions
We can find the derivative of the hyperbolic function by using the definition. By
transforming the hyperbolic function into the linear combination of ex and e−x , we
can differentiate the resulting equation.
Example 8 By using the definition, show that
a)
d
sinh x = cosh x.
dx
b)
d
cosh x = sinh x.
dx
We can show that for hyperbolic function the differentiation are given by,
u(x)
sinh u
cosh u
tanh u
cosech u
sech u
coth u
u0 (x)
du
dx
du
sinh u
dx
du
sech2 u
dx
cosh u
du
dx
du
− sech u tanh u
dx
du
− cosech2 u
dx
− cosech u coth u
Example 9: Find the derivative of the following functions
a) y = sinh(2x − 5).
b) y = sinh3 (2x − 5).
p
c) y = sinh (2x − 5).
p
d) y = sinh3 (2x − 5).
e) y = cosh(x2 − 3).
f) y = cosh2 (x2 − 3).
g) y = tanh4 (3x − 2).
h) y = tanh
x
.
x−1
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i) y =
ex
.
sinh 2x
j) y = ln(tanh x2 ).
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Differentiation Involving Inverse Trigonometric
Functions
Refer to the formula for these differentiation. Make sure that when you use it, do the
chain rule as well for the composite function. In fact it would be easier for you to
refer every inverse trigonometric function as a composite function, and do the chain
rule when applying the formula to find the derivative.
Example 10 : Find the derivative of the following functions
a) y = sin-1 6x.
1 + 4x
-1
b) y = tan
.
1 − 4x
c) y = sec-1 e3x .
d) y = cos-1 x2 .
e) y = cot-1
√
x.
f) y = cosec-1 (x2 + 1).
g) y = sec-1 (2x + 1).
h) y = tan-1 (ln x).
i) y = x sin-1 x +
8
√
1 − x2 .
Differentiation Involving Inverse Hyperbolic Functions
Same as the inverse trigonometric, you will just need to refer to the formula for the
differentiation of the inverse hyperbolic function.
Example 11 : Find the derivative of the following functions
a) y = sinh-1 (3x + 2).
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b) y = cosh-1
√
x2 + 1.
c) y = (1 − x) tanh-1 x.
d) y = (1 − x) coth-1
√
x.
e) y = cos-1 x − x sech-1 x.
f) y = cosh-1 (sec x).
g) y = x sinh-1 x1 .
h) y = ex sech-1 x.
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Exercise
1. Differentiate the following functions with respect to x
a) ex tan x.
cot 2x
b)
.
x
2. Find the derivative of ln(ax + b).
√
(1 + 2x) 1 + x
3. By using logarithmic differentiation, find the derivative of
.
1−x
4. Given y = ex cos x. Show that
dy
d2 y
− 4 + 5y = 0.
2
dx
dx
√
5. Differentiate cos-1 (2x 1 − x2 ).
6. Differentiate x2 (sin-1 x)3 .
7. Find the derivative of sin-1 ex + 2 tan-1 (3x).
√
8. Given y = 1 − x2 sin x, show that
(1 − x2 )
9. Find
dy
+ xy = 1 − x2 .
dx
dy
for the following functions by implicit differentiation
dx
a) tan-1 y = sin-1 x.
b) sin-1 (xy) = cos-1 (x − y).
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10
Conclusion
In this chapter, you should be able to :
1. Find the derivatives of functions involving products and quotient.
2. Using chain rule.
3. Use logarithmic differentiation.
4. Find the derivatives of hyperbolic functions implicitly and explicitly.
5. Know how to properly read the formula.
6. Find the derivatives of inverse trigonometric functions by using the formula
combined with chain rule.
7. Find the derivatives of inverse hyperbolic functions by using the formula combined with chain rule.
8. Most importantly, recognised straight away the technique of differentiation that
is necessary to find the derivation of a given function.
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