Chapter
stringclasses 18
values | sentence_range
stringlengths 3
9
| Text
stringlengths 7
7.34k
|
|---|---|---|
1 | 2018-2021 | The derivative of f at c is
defined by
0
(
)
( )
lim
h
f c
h
f c
h
→
+
−
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
119
f (x)
xn
sin x
cos x
tan x
f ′(x)
nxn – 1
cos x
– sin x
sec2 x
provided this limit exists Derivative of f at c is denoted by f ′(c) or
( ( )) |c
d
dxf x The
function defined by
0
(
)
( )
( )
lim
h
f x
h
f x
f
x
h
→
+
−
′
=
wherever the limit exists is defined to be the derivative of f The derivative of f is
denoted by f ′(x) or
d( ( ))
dxf x
or if y = f (x) by dy
dx or y′ |
1 | 2019-2022 | Derivative of f at c is denoted by f ′(c) or
( ( )) |c
d
dxf x The
function defined by
0
(
)
( )
( )
lim
h
f x
h
f x
f
x
h
→
+
−
′
=
wherever the limit exists is defined to be the derivative of f The derivative of f is
denoted by f ′(x) or
d( ( ))
dxf x
or if y = f (x) by dy
dx or y′ The process of finding
derivative of a function is called differentiation |
1 | 2020-2023 | The
function defined by
0
(
)
( )
( )
lim
h
f x
h
f x
f
x
h
→
+
−
′
=
wherever the limit exists is defined to be the derivative of f The derivative of f is
denoted by f ′(x) or
d( ( ))
dxf x
or if y = f (x) by dy
dx or y′ The process of finding
derivative of a function is called differentiation We also use the phrase differentiate
f(x) with respect to x to mean find f ′(x) |
1 | 2021-2024 | The derivative of f is
denoted by f ′(x) or
d( ( ))
dxf x
or if y = f (x) by dy
dx or y′ The process of finding
derivative of a function is called differentiation We also use the phrase differentiate
f(x) with respect to x to mean find f ′(x) (1)The following rules were established as a part of algebra of derivatives:
(u ± v)′ = u′ ± v′
(2)
(uv)′ = u′v + uv′ (Leibnitz or product rule)
(3)
2
u
u v
uv
v
v
′
′ −
′
=
, wherever v ≠ 0 (Quotient rule) |
1 | 2022-2025 | The process of finding
derivative of a function is called differentiation We also use the phrase differentiate
f(x) with respect to x to mean find f ′(x) (1)The following rules were established as a part of algebra of derivatives:
(u ± v)′ = u′ ± v′
(2)
(uv)′ = u′v + uv′ (Leibnitz or product rule)
(3)
2
u
u v
uv
v
v
′
′ −
′
=
, wherever v ≠ 0 (Quotient rule) The following table gives a list of derivatives of certain standard functions:
Table 5 |
1 | 2023-2026 | We also use the phrase differentiate
f(x) with respect to x to mean find f ′(x) (1)The following rules were established as a part of algebra of derivatives:
(u ± v)′ = u′ ± v′
(2)
(uv)′ = u′v + uv′ (Leibnitz or product rule)
(3)
2
u
u v
uv
v
v
′
′ −
′
=
, wherever v ≠ 0 (Quotient rule) The following table gives a list of derivatives of certain standard functions:
Table 5 3
Whenever we defined derivative, we had put a caution provided the limit exists |
1 | 2024-2027 | (1)The following rules were established as a part of algebra of derivatives:
(u ± v)′ = u′ ± v′
(2)
(uv)′ = u′v + uv′ (Leibnitz or product rule)
(3)
2
u
u v
uv
v
v
′
′ −
′
=
, wherever v ≠ 0 (Quotient rule) The following table gives a list of derivatives of certain standard functions:
Table 5 3
Whenever we defined derivative, we had put a caution provided the limit exists Now the natural question is; what if it doesn’t |
1 | 2025-2028 | The following table gives a list of derivatives of certain standard functions:
Table 5 3
Whenever we defined derivative, we had put a caution provided the limit exists Now the natural question is; what if it doesn’t The question is quite pertinent and so is
its answer |
1 | 2026-2029 | 3
Whenever we defined derivative, we had put a caution provided the limit exists Now the natural question is; what if it doesn’t The question is quite pertinent and so is
its answer If
0
(
)
( )
lim
h
f c
h
f c
h
→
+
−
does not exist, we say that f is not differentiable at c |
1 | 2027-2030 | Now the natural question is; what if it doesn’t The question is quite pertinent and so is
its answer If
0
(
)
( )
lim
h
f c
h
f c
h
→
+
−
does not exist, we say that f is not differentiable at c In other words, we say that a function f is differentiable at a point c in its domain if both
–
0
(
)
( )
lim
h
f c
h
f c
h
→
+
−
and
0
(
)
( )
lim
h
f c
h
f c
h
+
→
+
−
are finite and equal |
1 | 2028-2031 | The question is quite pertinent and so is
its answer If
0
(
)
( )
lim
h
f c
h
f c
h
→
+
−
does not exist, we say that f is not differentiable at c In other words, we say that a function f is differentiable at a point c in its domain if both
–
0
(
)
( )
lim
h
f c
h
f c
h
→
+
−
and
0
(
)
( )
lim
h
f c
h
f c
h
+
→
+
−
are finite and equal A function is said
to be differentiable in an interval [a, b] if it is differentiable at every point of [a, b] |
1 | 2029-2032 | If
0
(
)
( )
lim
h
f c
h
f c
h
→
+
−
does not exist, we say that f is not differentiable at c In other words, we say that a function f is differentiable at a point c in its domain if both
–
0
(
)
( )
lim
h
f c
h
f c
h
→
+
−
and
0
(
)
( )
lim
h
f c
h
f c
h
+
→
+
−
are finite and equal A function is said
to be differentiable in an interval [a, b] if it is differentiable at every point of [a, b] As
in case of continuity, at the end points a and b, we take the right hand limit and left hand
limit, which are nothing but left hand derivative and right hand derivative of the function
at a and b respectively |
1 | 2030-2033 | In other words, we say that a function f is differentiable at a point c in its domain if both
–
0
(
)
( )
lim
h
f c
h
f c
h
→
+
−
and
0
(
)
( )
lim
h
f c
h
f c
h
+
→
+
−
are finite and equal A function is said
to be differentiable in an interval [a, b] if it is differentiable at every point of [a, b] As
in case of continuity, at the end points a and b, we take the right hand limit and left hand
limit, which are nothing but left hand derivative and right hand derivative of the function
at a and b respectively Similarly, a function is said to be differentiable in an interval
(a, b) if it is differentiable at every point of (a, b) |
1 | 2031-2034 | A function is said
to be differentiable in an interval [a, b] if it is differentiable at every point of [a, b] As
in case of continuity, at the end points a and b, we take the right hand limit and left hand
limit, which are nothing but left hand derivative and right hand derivative of the function
at a and b respectively Similarly, a function is said to be differentiable in an interval
(a, b) if it is differentiable at every point of (a, b) Rationalised 2023-24
MATHEMATICS
120
Theorem 3 If a function f is differentiable at a point c, then it is also continuous at that
point |
1 | 2032-2035 | As
in case of continuity, at the end points a and b, we take the right hand limit and left hand
limit, which are nothing but left hand derivative and right hand derivative of the function
at a and b respectively Similarly, a function is said to be differentiable in an interval
(a, b) if it is differentiable at every point of (a, b) Rationalised 2023-24
MATHEMATICS
120
Theorem 3 If a function f is differentiable at a point c, then it is also continuous at that
point Proof Since f is differentiable at c, we have
( )
( )
lim
( )
x
c
f x
f c
f c
x
c
→
−
=
′
−
But for x ≠ c, we have
f (x) – f (c) =
( )
( ) |
1 | 2033-2036 | Similarly, a function is said to be differentiable in an interval
(a, b) if it is differentiable at every point of (a, b) Rationalised 2023-24
MATHEMATICS
120
Theorem 3 If a function f is differentiable at a point c, then it is also continuous at that
point Proof Since f is differentiable at c, we have
( )
( )
lim
( )
x
c
f x
f c
f c
x
c
→
−
=
′
−
But for x ≠ c, we have
f (x) – f (c) =
( )
( ) (
)
f x
f c
x
c
x
c
−
−
−
Therefore
lim[ ( )
( )]
x
c f x
f c
→
−
=
( )
( )
lim |
1 | 2034-2037 | Rationalised 2023-24
MATHEMATICS
120
Theorem 3 If a function f is differentiable at a point c, then it is also continuous at that
point Proof Since f is differentiable at c, we have
( )
( )
lim
( )
x
c
f x
f c
f c
x
c
→
−
=
′
−
But for x ≠ c, we have
f (x) – f (c) =
( )
( ) (
)
f x
f c
x
c
x
c
−
−
−
Therefore
lim[ ( )
( )]
x
c f x
f c
→
−
=
( )
( )
lim (
)
x
c
f x
f c
x
c
x
c
→
−
−
−
or
lim[ ( )]
lim[ ( )]
x
c
x
c
f x
f c
→
−→
=
( )
( )
lim |
1 | 2035-2038 | Proof Since f is differentiable at c, we have
( )
( )
lim
( )
x
c
f x
f c
f c
x
c
→
−
=
′
−
But for x ≠ c, we have
f (x) – f (c) =
( )
( ) (
)
f x
f c
x
c
x
c
−
−
−
Therefore
lim[ ( )
( )]
x
c f x
f c
→
−
=
( )
( )
lim (
)
x
c
f x
f c
x
c
x
c
→
−
−
−
or
lim[ ( )]
lim[ ( )]
x
c
x
c
f x
f c
→
−→
=
( )
( )
lim lim [(
)]
x
c
x
c
f x
f c
x
c
x
c
→
→
−
−
−
= f′(c) |
1 | 2036-2039 | (
)
f x
f c
x
c
x
c
−
−
−
Therefore
lim[ ( )
( )]
x
c f x
f c
→
−
=
( )
( )
lim (
)
x
c
f x
f c
x
c
x
c
→
−
−
−
or
lim[ ( )]
lim[ ( )]
x
c
x
c
f x
f c
→
−→
=
( )
( )
lim lim [(
)]
x
c
x
c
f x
f c
x
c
x
c
→
→
−
−
−
= f′(c) 0 = 0
or
lim
( )
x
c f x
→
= f (c)
Hence f is continuous at x = c |
1 | 2037-2040 | (
)
x
c
f x
f c
x
c
x
c
→
−
−
−
or
lim[ ( )]
lim[ ( )]
x
c
x
c
f x
f c
→
−→
=
( )
( )
lim lim [(
)]
x
c
x
c
f x
f c
x
c
x
c
→
→
−
−
−
= f′(c) 0 = 0
or
lim
( )
x
c f x
→
= f (c)
Hence f is continuous at x = c Corollary 1 Every differentiable function is continuous |
1 | 2038-2041 | lim [(
)]
x
c
x
c
f x
f c
x
c
x
c
→
→
−
−
−
= f′(c) 0 = 0
or
lim
( )
x
c f x
→
= f (c)
Hence f is continuous at x = c Corollary 1 Every differentiable function is continuous We remark that the converse of the above statement is not true |
1 | 2039-2042 | 0 = 0
or
lim
( )
x
c f x
→
= f (c)
Hence f is continuous at x = c Corollary 1 Every differentiable function is continuous We remark that the converse of the above statement is not true Indeed we have
seen that the function defined by f(x) = |x | is a continuous function |
1 | 2040-2043 | Corollary 1 Every differentiable function is continuous We remark that the converse of the above statement is not true Indeed we have
seen that the function defined by f(x) = |x | is a continuous function Consider the left
hand limit
0–
(0
)
(0)
lim
1
h
f
h
f
h
h
h
→
+
−
=−
= −
The right hand limit
0
(0
)
(0)
lim
1
h
f
h
f
h
h
h
+
→
+
−
=
=
Since the above left and right hand limits at 0 are not equal,
0
(0
)
(0)
lim
h
f
h
f
h
→
+
−
does not exist and hence f is not differentiable at 0 |
1 | 2041-2044 | We remark that the converse of the above statement is not true Indeed we have
seen that the function defined by f(x) = |x | is a continuous function Consider the left
hand limit
0–
(0
)
(0)
lim
1
h
f
h
f
h
h
h
→
+
−
=−
= −
The right hand limit
0
(0
)
(0)
lim
1
h
f
h
f
h
h
h
+
→
+
−
=
=
Since the above left and right hand limits at 0 are not equal,
0
(0
)
(0)
lim
h
f
h
f
h
→
+
−
does not exist and hence f is not differentiable at 0 Thus f is not a differentiable
function |
1 | 2042-2045 | Indeed we have
seen that the function defined by f(x) = |x | is a continuous function Consider the left
hand limit
0–
(0
)
(0)
lim
1
h
f
h
f
h
h
h
→
+
−
=−
= −
The right hand limit
0
(0
)
(0)
lim
1
h
f
h
f
h
h
h
+
→
+
−
=
=
Since the above left and right hand limits at 0 are not equal,
0
(0
)
(0)
lim
h
f
h
f
h
→
+
−
does not exist and hence f is not differentiable at 0 Thus f is not a differentiable
function 5 |
1 | 2043-2046 | Consider the left
hand limit
0–
(0
)
(0)
lim
1
h
f
h
f
h
h
h
→
+
−
=−
= −
The right hand limit
0
(0
)
(0)
lim
1
h
f
h
f
h
h
h
+
→
+
−
=
=
Since the above left and right hand limits at 0 are not equal,
0
(0
)
(0)
lim
h
f
h
f
h
→
+
−
does not exist and hence f is not differentiable at 0 Thus f is not a differentiable
function 5 3 |
1 | 2044-2047 | Thus f is not a differentiable
function 5 3 1 Derivatives of composite functions
To study derivative of composite functions, we start with an illustrative example |
1 | 2045-2048 | 5 3 1 Derivatives of composite functions
To study derivative of composite functions, we start with an illustrative example Say,
we want to find the derivative of f, where
f (x) = (2x + 1)3
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
121
One way is to expand (2x + 1)3 using binomial theorem and find the derivative as
a polynomial function as illustrated below |
1 | 2046-2049 | 3 1 Derivatives of composite functions
To study derivative of composite functions, we start with an illustrative example Say,
we want to find the derivative of f, where
f (x) = (2x + 1)3
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
121
One way is to expand (2x + 1)3 using binomial theorem and find the derivative as
a polynomial function as illustrated below ( )
dxd f x
=
3
(2
1)
d
x
dx
+
=
3
2
(8
12
6
1)
d
x
x
x
dx
+
+
+
= 24x2 + 24x + 6
= 6 (2x + 1)2
Now, observe that
f (x) = (h o g) (x)
where g(x) = 2x + 1 and h(x) = x3 |
1 | 2047-2050 | 1 Derivatives of composite functions
To study derivative of composite functions, we start with an illustrative example Say,
we want to find the derivative of f, where
f (x) = (2x + 1)3
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
121
One way is to expand (2x + 1)3 using binomial theorem and find the derivative as
a polynomial function as illustrated below ( )
dxd f x
=
3
(2
1)
d
x
dx
+
=
3
2
(8
12
6
1)
d
x
x
x
dx
+
+
+
= 24x2 + 24x + 6
= 6 (2x + 1)2
Now, observe that
f (x) = (h o g) (x)
where g(x) = 2x + 1 and h(x) = x3 Put t = g(x) = 2x + 1 |
1 | 2048-2051 | Say,
we want to find the derivative of f, where
f (x) = (2x + 1)3
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
121
One way is to expand (2x + 1)3 using binomial theorem and find the derivative as
a polynomial function as illustrated below ( )
dxd f x
=
3
(2
1)
d
x
dx
+
=
3
2
(8
12
6
1)
d
x
x
x
dx
+
+
+
= 24x2 + 24x + 6
= 6 (2x + 1)2
Now, observe that
f (x) = (h o g) (x)
where g(x) = 2x + 1 and h(x) = x3 Put t = g(x) = 2x + 1 Then f(x) = h(t) = t3 |
1 | 2049-2052 | ( )
dxd f x
=
3
(2
1)
d
x
dx
+
=
3
2
(8
12
6
1)
d
x
x
x
dx
+
+
+
= 24x2 + 24x + 6
= 6 (2x + 1)2
Now, observe that
f (x) = (h o g) (x)
where g(x) = 2x + 1 and h(x) = x3 Put t = g(x) = 2x + 1 Then f(x) = h(t) = t3 Thus
df
dx =
6 (2x + 1)2 = 3(2x + 1)2 |
1 | 2050-2053 | Put t = g(x) = 2x + 1 Then f(x) = h(t) = t3 Thus
df
dx =
6 (2x + 1)2 = 3(2x + 1)2 2 = 3t2 |
1 | 2051-2054 | Then f(x) = h(t) = t3 Thus
df
dx =
6 (2x + 1)2 = 3(2x + 1)2 2 = 3t2 2 = dh dt
dt
dx
⋅
The advantage with such observation is that it simplifies the calculation in finding
the derivative of, say, (2x + 1)100 |
1 | 2052-2055 | Thus
df
dx =
6 (2x + 1)2 = 3(2x + 1)2 2 = 3t2 2 = dh dt
dt
dx
⋅
The advantage with such observation is that it simplifies the calculation in finding
the derivative of, say, (2x + 1)100 We may formalise this observation in the following
theorem called the chain rule |
1 | 2053-2056 | 2 = 3t2 2 = dh dt
dt
dx
⋅
The advantage with such observation is that it simplifies the calculation in finding
the derivative of, say, (2x + 1)100 We may formalise this observation in the following
theorem called the chain rule Theorem 4 (Chain Rule) Let f be a real valued function which is a composite of two
functions u and v; i |
1 | 2054-2057 | 2 = dh dt
dt
dx
⋅
The advantage with such observation is that it simplifies the calculation in finding
the derivative of, say, (2x + 1)100 We may formalise this observation in the following
theorem called the chain rule Theorem 4 (Chain Rule) Let f be a real valued function which is a composite of two
functions u and v; i e |
1 | 2055-2058 | We may formalise this observation in the following
theorem called the chain rule Theorem 4 (Chain Rule) Let f be a real valued function which is a composite of two
functions u and v; i e , f = v o u |
1 | 2056-2059 | Theorem 4 (Chain Rule) Let f be a real valued function which is a composite of two
functions u and v; i e , f = v o u Suppose t = u(x) and if both dt
dx and dv
dt exist, we have
df
dv dt
dx
=dt dx
⋅
We skip the proof of this theorem |
1 | 2057-2060 | e , f = v o u Suppose t = u(x) and if both dt
dx and dv
dt exist, we have
df
dv dt
dx
=dt dx
⋅
We skip the proof of this theorem Chain rule may be extended as follows |
1 | 2058-2061 | , f = v o u Suppose t = u(x) and if both dt
dx and dv
dt exist, we have
df
dv dt
dx
=dt dx
⋅
We skip the proof of this theorem Chain rule may be extended as follows Suppose
f is a real valued function which is a composite of three functions u, v and w; i |
1 | 2059-2062 | Suppose t = u(x) and if both dt
dx and dv
dt exist, we have
df
dv dt
dx
=dt dx
⋅
We skip the proof of this theorem Chain rule may be extended as follows Suppose
f is a real valued function which is a composite of three functions u, v and w; i e |
1 | 2060-2063 | Chain rule may be extended as follows Suppose
f is a real valued function which is a composite of three functions u, v and w; i e ,
f = (w o u) o v |
1 | 2061-2064 | Suppose
f is a real valued function which is a composite of three functions u, v and w; i e ,
f = (w o u) o v If t = v (x) and s = u (t), then
( o )
df
d w u
dt
dw ds
dt
dx
dt
dx
ds
dt
dx
=
⋅
=
⋅
⋅
provided all the derivatives in the statement exist |
1 | 2062-2065 | e ,
f = (w o u) o v If t = v (x) and s = u (t), then
( o )
df
d w u
dt
dw ds
dt
dx
dt
dx
ds
dt
dx
=
⋅
=
⋅
⋅
provided all the derivatives in the statement exist Reader is invited to formulate chain
rule for composite of more functions |
1 | 2063-2066 | ,
f = (w o u) o v If t = v (x) and s = u (t), then
( o )
df
d w u
dt
dw ds
dt
dx
dt
dx
ds
dt
dx
=
⋅
=
⋅
⋅
provided all the derivatives in the statement exist Reader is invited to formulate chain
rule for composite of more functions Example 21 Find the derivative of the function given by f(x) = sin (x2) |
1 | 2064-2067 | If t = v (x) and s = u (t), then
( o )
df
d w u
dt
dw ds
dt
dx
dt
dx
ds
dt
dx
=
⋅
=
⋅
⋅
provided all the derivatives in the statement exist Reader is invited to formulate chain
rule for composite of more functions Example 21 Find the derivative of the function given by f(x) = sin (x2) Solution Observe that the given function is a composite of two functions |
1 | 2065-2068 | Reader is invited to formulate chain
rule for composite of more functions Example 21 Find the derivative of the function given by f(x) = sin (x2) Solution Observe that the given function is a composite of two functions Indeed, if
t = u(x) = x2 and v(t) = sin t, then
f (x) = (v o u) (x) = v(u(x)) = v(x2) = sin x2
Rationalised 2023-24
MATHEMATICS
122
Put t = u(x) = x2 |
1 | 2066-2069 | Example 21 Find the derivative of the function given by f(x) = sin (x2) Solution Observe that the given function is a composite of two functions Indeed, if
t = u(x) = x2 and v(t) = sin t, then
f (x) = (v o u) (x) = v(u(x)) = v(x2) = sin x2
Rationalised 2023-24
MATHEMATICS
122
Put t = u(x) = x2 Observe that
cos
dv
t
dt =
and
2
dt
dx =x
exist |
1 | 2067-2070 | Solution Observe that the given function is a composite of two functions Indeed, if
t = u(x) = x2 and v(t) = sin t, then
f (x) = (v o u) (x) = v(u(x)) = v(x2) = sin x2
Rationalised 2023-24
MATHEMATICS
122
Put t = u(x) = x2 Observe that
cos
dv
t
dt =
and
2
dt
dx =x
exist Hence, by chain rule
df
dx =
cos
2
dv
dt
t
x
dt
⋅dx
=
⋅
It is normal practice to express the final result only in terms of x |
1 | 2068-2071 | Indeed, if
t = u(x) = x2 and v(t) = sin t, then
f (x) = (v o u) (x) = v(u(x)) = v(x2) = sin x2
Rationalised 2023-24
MATHEMATICS
122
Put t = u(x) = x2 Observe that
cos
dv
t
dt =
and
2
dt
dx =x
exist Hence, by chain rule
df
dx =
cos
2
dv
dt
t
x
dt
⋅dx
=
⋅
It is normal practice to express the final result only in terms of x Thus
df
dx =
2
cos
2
2 cos
t
x
x
x
⋅
=
EXERCISE 5 |
1 | 2069-2072 | Observe that
cos
dv
t
dt =
and
2
dt
dx =x
exist Hence, by chain rule
df
dx =
cos
2
dv
dt
t
x
dt
⋅dx
=
⋅
It is normal practice to express the final result only in terms of x Thus
df
dx =
2
cos
2
2 cos
t
x
x
x
⋅
=
EXERCISE 5 2
Differentiate the functions with respect to x in Exercises 1 to 8 |
1 | 2070-2073 | Hence, by chain rule
df
dx =
cos
2
dv
dt
t
x
dt
⋅dx
=
⋅
It is normal practice to express the final result only in terms of x Thus
df
dx =
2
cos
2
2 cos
t
x
x
x
⋅
=
EXERCISE 5 2
Differentiate the functions with respect to x in Exercises 1 to 8 1 |
1 | 2071-2074 | Thus
df
dx =
2
cos
2
2 cos
t
x
x
x
⋅
=
EXERCISE 5 2
Differentiate the functions with respect to x in Exercises 1 to 8 1 sin (x2 + 5)
2 |
1 | 2072-2075 | 2
Differentiate the functions with respect to x in Exercises 1 to 8 1 sin (x2 + 5)
2 cos (sin x)
3 |
1 | 2073-2076 | 1 sin (x2 + 5)
2 cos (sin x)
3 sin (ax + b)
4 |
1 | 2074-2077 | sin (x2 + 5)
2 cos (sin x)
3 sin (ax + b)
4 sec (tan (
x ))
5 |
1 | 2075-2078 | cos (sin x)
3 sin (ax + b)
4 sec (tan (
x ))
5 sin (
)
cos (
)
ax
b
cx
d
++
6 |
1 | 2076-2079 | sin (ax + b)
4 sec (tan (
x ))
5 sin (
)
cos (
)
ax
b
cx
d
++
6 cos x3 |
1 | 2077-2080 | sec (tan (
x ))
5 sin (
)
cos (
)
ax
b
cx
d
++
6 cos x3 sin2 (x5)
7 |
1 | 2078-2081 | sin (
)
cos (
)
ax
b
cx
d
++
6 cos x3 sin2 (x5)
7 (
2 cot x2)
8 |
1 | 2079-2082 | cos x3 sin2 (x5)
7 (
2 cot x2)
8 (
)
cos
x
9 |
1 | 2080-2083 | sin2 (x5)
7 (
2 cot x2)
8 (
)
cos
x
9 Prove that the function f given by
f (x) = |x – 1|, x ∈ R
is not differentiable at x = 1 |
1 | 2081-2084 | (
2 cot x2)
8 (
)
cos
x
9 Prove that the function f given by
f (x) = |x – 1|, x ∈ R
is not differentiable at x = 1 10 |
1 | 2082-2085 | (
)
cos
x
9 Prove that the function f given by
f (x) = |x – 1|, x ∈ R
is not differentiable at x = 1 10 Prove that the greatest integer function defined by
f (x) = [x], 0 < x < 3
is not differentiable at x = 1 and x = 2 |
1 | 2083-2086 | Prove that the function f given by
f (x) = |x – 1|, x ∈ R
is not differentiable at x = 1 10 Prove that the greatest integer function defined by
f (x) = [x], 0 < x < 3
is not differentiable at x = 1 and x = 2 5 |
1 | 2084-2087 | 10 Prove that the greatest integer function defined by
f (x) = [x], 0 < x < 3
is not differentiable at x = 1 and x = 2 5 3 |
1 | 2085-2088 | Prove that the greatest integer function defined by
f (x) = [x], 0 < x < 3
is not differentiable at x = 1 and x = 2 5 3 2 Derivatives of implicit functions
Until now we have been differentiating various functions given in the form y = f(x) |
1 | 2086-2089 | 5 3 2 Derivatives of implicit functions
Until now we have been differentiating various functions given in the form y = f(x) But it is not necessary that functions are always expressed in this form |
1 | 2087-2090 | 3 2 Derivatives of implicit functions
Until now we have been differentiating various functions given in the form y = f(x) But it is not necessary that functions are always expressed in this form For example,
consider one of the following relationships between x and y:
x – y – π = 0
x + sin xy – y = 0
In the first case, we can solve for y and rewrite the relationship as y = x – π |
1 | 2088-2091 | 2 Derivatives of implicit functions
Until now we have been differentiating various functions given in the form y = f(x) But it is not necessary that functions are always expressed in this form For example,
consider one of the following relationships between x and y:
x – y – π = 0
x + sin xy – y = 0
In the first case, we can solve for y and rewrite the relationship as y = x – π In
the second case, it does not seem that there is an easy way to solve for y |
1 | 2089-2092 | But it is not necessary that functions are always expressed in this form For example,
consider one of the following relationships between x and y:
x – y – π = 0
x + sin xy – y = 0
In the first case, we can solve for y and rewrite the relationship as y = x – π In
the second case, it does not seem that there is an easy way to solve for y Nevertheless,
there is no doubt about the dependence of y on x in either of the cases |
1 | 2090-2093 | For example,
consider one of the following relationships between x and y:
x – y – π = 0
x + sin xy – y = 0
In the first case, we can solve for y and rewrite the relationship as y = x – π In
the second case, it does not seem that there is an easy way to solve for y Nevertheless,
there is no doubt about the dependence of y on x in either of the cases When a
relationship between x and y is expressed in a way that it is easy to solve for y and
write y = f(x), we say that y is given as an explicit function of x |
1 | 2091-2094 | In
the second case, it does not seem that there is an easy way to solve for y Nevertheless,
there is no doubt about the dependence of y on x in either of the cases When a
relationship between x and y is expressed in a way that it is easy to solve for y and
write y = f(x), we say that y is given as an explicit function of x In the latter case it
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
123
is implicit that y is a function of x and we say that the relationship of the second type,
above, gives function implicitly |
1 | 2092-2095 | Nevertheless,
there is no doubt about the dependence of y on x in either of the cases When a
relationship between x and y is expressed in a way that it is easy to solve for y and
write y = f(x), we say that y is given as an explicit function of x In the latter case it
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
123
is implicit that y is a function of x and we say that the relationship of the second type,
above, gives function implicitly In this subsection, we learn to differentiate implicit
functions |
1 | 2093-2096 | When a
relationship between x and y is expressed in a way that it is easy to solve for y and
write y = f(x), we say that y is given as an explicit function of x In the latter case it
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
123
is implicit that y is a function of x and we say that the relationship of the second type,
above, gives function implicitly In this subsection, we learn to differentiate implicit
functions Example 22 Find dy
dx if x – y = π |
1 | 2094-2097 | In the latter case it
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
123
is implicit that y is a function of x and we say that the relationship of the second type,
above, gives function implicitly In this subsection, we learn to differentiate implicit
functions Example 22 Find dy
dx if x – y = π Solution One way is to solve for y and rewrite the above as
y = x – π
But then
dy
dx = 1
Alternatively, directly differentiating the relationship w |
1 | 2095-2098 | In this subsection, we learn to differentiate implicit
functions Example 22 Find dy
dx if x – y = π Solution One way is to solve for y and rewrite the above as
y = x – π
But then
dy
dx = 1
Alternatively, directly differentiating the relationship w r |
1 | 2096-2099 | Example 22 Find dy
dx if x – y = π Solution One way is to solve for y and rewrite the above as
y = x – π
But then
dy
dx = 1
Alternatively, directly differentiating the relationship w r t |
1 | 2097-2100 | Solution One way is to solve for y and rewrite the above as
y = x – π
But then
dy
dx = 1
Alternatively, directly differentiating the relationship w r t , x, we have
(
)
d
x
y
dx
−
= d
dx
π
Recall that d
dx
π means to differentiate the constant function taking value π
everywhere w |
1 | 2098-2101 | r t , x, we have
(
)
d
x
y
dx
−
= d
dx
π
Recall that d
dx
π means to differentiate the constant function taking value π
everywhere w r |
1 | 2099-2102 | t , x, we have
(
)
d
x
y
dx
−
= d
dx
π
Recall that d
dx
π means to differentiate the constant function taking value π
everywhere w r t |
1 | 2100-2103 | , x, we have
(
)
d
x
y
dx
−
= d
dx
π
Recall that d
dx
π means to differentiate the constant function taking value π
everywhere w r t , x |
1 | 2101-2104 | r t , x Thus
( )
( )
d
d
x
y
dx
−dx
= 0
which implies that
dy
dx =
1
dx
dx =
Example 23 Find dy
dx
, if y + sin y = cos x |
1 | 2102-2105 | t , x Thus
( )
( )
d
d
x
y
dx
−dx
= 0
which implies that
dy
dx =
1
dx
dx =
Example 23 Find dy
dx
, if y + sin y = cos x Solution We differentiate the relationship directly with respect to x, i |
1 | 2103-2106 | , x Thus
( )
( )
d
d
x
y
dx
−dx
= 0
which implies that
dy
dx =
1
dx
dx =
Example 23 Find dy
dx
, if y + sin y = cos x Solution We differentiate the relationship directly with respect to x, i e |
1 | 2104-2107 | Thus
( )
( )
d
d
x
y
dx
−dx
= 0
which implies that
dy
dx =
1
dx
dx =
Example 23 Find dy
dx
, if y + sin y = cos x Solution We differentiate the relationship directly with respect to x, i e ,
(sin )
dy
d
y
dx
+dx
=
d(cos )
x
dx
which implies using chain rule
cos
dy
ydy
dx
dx
+
⋅
= – sin x
This gives
dy
dx =
sin
1
cos
x
y
− +
where
y ≠ (2n + 1) π
Rationalised 2023-24
MATHEMATICS
124
5 |
1 | 2105-2108 | Solution We differentiate the relationship directly with respect to x, i e ,
(sin )
dy
d
y
dx
+dx
=
d(cos )
x
dx
which implies using chain rule
cos
dy
ydy
dx
dx
+
⋅
= – sin x
This gives
dy
dx =
sin
1
cos
x
y
− +
where
y ≠ (2n + 1) π
Rationalised 2023-24
MATHEMATICS
124
5 3 |
1 | 2106-2109 | e ,
(sin )
dy
d
y
dx
+dx
=
d(cos )
x
dx
which implies using chain rule
cos
dy
ydy
dx
dx
+
⋅
= – sin x
This gives
dy
dx =
sin
1
cos
x
y
− +
where
y ≠ (2n + 1) π
Rationalised 2023-24
MATHEMATICS
124
5 3 3 Derivatives of inverse trigonometric functions
We remark that inverse trigonometric functions are continuous functions, but we will
not prove this |
1 | 2107-2110 | ,
(sin )
dy
d
y
dx
+dx
=
d(cos )
x
dx
which implies using chain rule
cos
dy
ydy
dx
dx
+
⋅
= – sin x
This gives
dy
dx =
sin
1
cos
x
y
− +
where
y ≠ (2n + 1) π
Rationalised 2023-24
MATHEMATICS
124
5 3 3 Derivatives of inverse trigonometric functions
We remark that inverse trigonometric functions are continuous functions, but we will
not prove this Now we use chain rule to find derivatives of these functions |
1 | 2108-2111 | 3 3 Derivatives of inverse trigonometric functions
We remark that inverse trigonometric functions are continuous functions, but we will
not prove this Now we use chain rule to find derivatives of these functions Example 24 Find the derivative of f given by f(x) = sin–1 x assuming it exists |
1 | 2109-2112 | 3 Derivatives of inverse trigonometric functions
We remark that inverse trigonometric functions are continuous functions, but we will
not prove this Now we use chain rule to find derivatives of these functions Example 24 Find the derivative of f given by f(x) = sin–1 x assuming it exists Solution Let y = sin–1 x |
1 | 2110-2113 | Now we use chain rule to find derivatives of these functions Example 24 Find the derivative of f given by f(x) = sin–1 x assuming it exists Solution Let y = sin–1 x Then, x = sin y |
1 | 2111-2114 | Example 24 Find the derivative of f given by f(x) = sin–1 x assuming it exists Solution Let y = sin–1 x Then, x = sin y Differentiating both sides w |
1 | 2112-2115 | Solution Let y = sin–1 x Then, x = sin y Differentiating both sides w r |
1 | 2113-2116 | Then, x = sin y Differentiating both sides w r t |
1 | 2114-2117 | Differentiating both sides w r t x, we get
1 = cos y dy
dx
which implies that
dy
dx =
1
1
1
cos
cos(sin
)
y
x
−
=
Observe that this is defined only for cos y ≠ 0, i |
1 | 2115-2118 | r t x, we get
1 = cos y dy
dx
which implies that
dy
dx =
1
1
1
cos
cos(sin
)
y
x
−
=
Observe that this is defined only for cos y ≠ 0, i e |
1 | 2116-2119 | t x, we get
1 = cos y dy
dx
which implies that
dy
dx =
1
1
1
cos
cos(sin
)
y
x
−
=
Observe that this is defined only for cos y ≠ 0, i e , sin–1 x ≠
,
2 2
−π π
, i |
1 | 2117-2120 | x, we get
1 = cos y dy
dx
which implies that
dy
dx =
1
1
1
cos
cos(sin
)
y
x
−
=
Observe that this is defined only for cos y ≠ 0, i e , sin–1 x ≠
,
2 2
−π π
, i e |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.