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1 | 2418-2421 | , (
2)
12
1
1
x
y
−
=
So
2
2
1
2
1
(1
) 2
(0
2 )
0
x
y y
y
x
−
+
−
=
Hence
(1 – x2) y2 – xy1 = 0
EXERCISE 5 7
Find the second order derivatives of the functions given in Exercises 1 to 10 1 |
1 | 2419-2422 | 2
(0
2 )
0
x
y y
y
x
−
+
−
=
Hence
(1 – x2) y2 – xy1 = 0
EXERCISE 5 7
Find the second order derivatives of the functions given in Exercises 1 to 10 1 x2 + 3x + 2
2 |
1 | 2420-2423 | 7
Find the second order derivatives of the functions given in Exercises 1 to 10 1 x2 + 3x + 2
2 x20
3 |
1 | 2421-2424 | 1 x2 + 3x + 2
2 x20
3 x |
1 | 2422-2425 | x2 + 3x + 2
2 x20
3 x cos x
4 |
1 | 2423-2426 | x20
3 x cos x
4 log x
5 |
1 | 2424-2427 | x cos x
4 log x
5 x3 log x
6 |
1 | 2425-2428 | cos x
4 log x
5 x3 log x
6 ex sin 5x
7 |
1 | 2426-2429 | log x
5 x3 log x
6 ex sin 5x
7 e6x cos 3x
8 |
1 | 2427-2430 | x3 log x
6 ex sin 5x
7 e6x cos 3x
8 tan–1 x
9 |
1 | 2428-2431 | ex sin 5x
7 e6x cos 3x
8 tan–1 x
9 log (log x)
10 |
1 | 2429-2432 | e6x cos 3x
8 tan–1 x
9 log (log x)
10 sin (log x)
11 |
1 | 2430-2433 | tan–1 x
9 log (log x)
10 sin (log x)
11 If y = 5 cos x – 3 sin x, prove that
2
2
0
d y
y
dx
+
=
Rationalised 2023-24
MATHEMATICS
140
12 |
1 | 2431-2434 | log (log x)
10 sin (log x)
11 If y = 5 cos x – 3 sin x, prove that
2
2
0
d y
y
dx
+
=
Rationalised 2023-24
MATHEMATICS
140
12 If y = cos–1 x, Find
2
2
d y
dx
in terms of y alone |
1 | 2432-2435 | sin (log x)
11 If y = 5 cos x – 3 sin x, prove that
2
2
0
d y
y
dx
+
=
Rationalised 2023-24
MATHEMATICS
140
12 If y = cos–1 x, Find
2
2
d y
dx
in terms of y alone 13 |
1 | 2433-2436 | If y = 5 cos x – 3 sin x, prove that
2
2
0
d y
y
dx
+
=
Rationalised 2023-24
MATHEMATICS
140
12 If y = cos–1 x, Find
2
2
d y
dx
in terms of y alone 13 If y = 3 cos (log x) + 4 sin (log x), show that x2 y2 + xy1 + y = 0
14 |
1 | 2434-2437 | If y = cos–1 x, Find
2
2
d y
dx
in terms of y alone 13 If y = 3 cos (log x) + 4 sin (log x), show that x2 y2 + xy1 + y = 0
14 If y = Aemx + Benx, show that
2
2
(
)
0
d y
dy
m
n
mny
dx
dx
−
+
+
=
15 |
1 | 2435-2438 | 13 If y = 3 cos (log x) + 4 sin (log x), show that x2 y2 + xy1 + y = 0
14 If y = Aemx + Benx, show that
2
2
(
)
0
d y
dy
m
n
mny
dx
dx
−
+
+
=
15 If y = 500e7x + 600e–7x, show that
2
2
49
d y
y
dx
=
16 |
1 | 2436-2439 | If y = 3 cos (log x) + 4 sin (log x), show that x2 y2 + xy1 + y = 0
14 If y = Aemx + Benx, show that
2
2
(
)
0
d y
dy
m
n
mny
dx
dx
−
+
+
=
15 If y = 500e7x + 600e–7x, show that
2
2
49
d y
y
dx
=
16 If ey
(x + 1) = 1, show that
2
2
2
d y
dy
dx
dx
=
17 |
1 | 2437-2440 | If y = Aemx + Benx, show that
2
2
(
)
0
d y
dy
m
n
mny
dx
dx
−
+
+
=
15 If y = 500e7x + 600e–7x, show that
2
2
49
d y
y
dx
=
16 If ey
(x + 1) = 1, show that
2
2
2
d y
dy
dx
dx
=
17 If y = (tan–1 x)2, show that (x2 + 1)2 y2 + 2x (x2 + 1) y1 = 2
Miscellaneous Examples
Example 39 Differentiate w |
1 | 2438-2441 | If y = 500e7x + 600e–7x, show that
2
2
49
d y
y
dx
=
16 If ey
(x + 1) = 1, show that
2
2
2
d y
dy
dx
dx
=
17 If y = (tan–1 x)2, show that (x2 + 1)2 y2 + 2x (x2 + 1) y1 = 2
Miscellaneous Examples
Example 39 Differentiate w r |
1 | 2439-2442 | If ey
(x + 1) = 1, show that
2
2
2
d y
dy
dx
dx
=
17 If y = (tan–1 x)2, show that (x2 + 1)2 y2 + 2x (x2 + 1) y1 = 2
Miscellaneous Examples
Example 39 Differentiate w r t |
1 | 2440-2443 | If y = (tan–1 x)2, show that (x2 + 1)2 y2 + 2x (x2 + 1) y1 = 2
Miscellaneous Examples
Example 39 Differentiate w r t x, the following function:
(i)
12
3
2
2
4
x
x
+
+
+
(ii) log7 (log x)
Solution
(i)
Let y =
12
3
2
2
4
x
x
+
+
+
=
1
1
2
2
2
(3
2)
(2
4)
x
x
−
+
+
+
Note that this function is defined at all real numbers
x > −32 |
1 | 2441-2444 | r t x, the following function:
(i)
12
3
2
2
4
x
x
+
+
+
(ii) log7 (log x)
Solution
(i)
Let y =
12
3
2
2
4
x
x
+
+
+
=
1
1
2
2
2
(3
2)
(2
4)
x
x
−
+
+
+
Note that this function is defined at all real numbers
x > −32 Therefore
dy
dx =
1
1
1
1
2
2
2
2
1
1
(3
2)
(3
2)
(2
4)
(2
4)
2
2
d
d
x
x
x
x
dx
dx
−
− −
+
⋅
+
+ −
+
⋅
+
= 1
2 3
2
3
21
2
4
4
21
2
23
(
)
( )
(
)
x
x
x
+
⋅
−
+
⋅
−
−
=
(
)
3
2
2
3
2
2 3
2
2
4
x
x
x
−
+
+
This is defined for all real numbers
x > −32 |
1 | 2442-2445 | t x, the following function:
(i)
12
3
2
2
4
x
x
+
+
+
(ii) log7 (log x)
Solution
(i)
Let y =
12
3
2
2
4
x
x
+
+
+
=
1
1
2
2
2
(3
2)
(2
4)
x
x
−
+
+
+
Note that this function is defined at all real numbers
x > −32 Therefore
dy
dx =
1
1
1
1
2
2
2
2
1
1
(3
2)
(3
2)
(2
4)
(2
4)
2
2
d
d
x
x
x
x
dx
dx
−
− −
+
⋅
+
+ −
+
⋅
+
= 1
2 3
2
3
21
2
4
4
21
2
23
(
)
( )
(
)
x
x
x
+
⋅
−
+
⋅
−
−
=
(
)
3
2
2
3
2
2 3
2
2
4
x
x
x
−
+
+
This is defined for all real numbers
x > −32 Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
141
(ii)
Let y = log7 (log x) = log (log )
log7
x (by change of base formula) |
1 | 2443-2446 | x, the following function:
(i)
12
3
2
2
4
x
x
+
+
+
(ii) log7 (log x)
Solution
(i)
Let y =
12
3
2
2
4
x
x
+
+
+
=
1
1
2
2
2
(3
2)
(2
4)
x
x
−
+
+
+
Note that this function is defined at all real numbers
x > −32 Therefore
dy
dx =
1
1
1
1
2
2
2
2
1
1
(3
2)
(3
2)
(2
4)
(2
4)
2
2
d
d
x
x
x
x
dx
dx
−
− −
+
⋅
+
+ −
+
⋅
+
= 1
2 3
2
3
21
2
4
4
21
2
23
(
)
( )
(
)
x
x
x
+
⋅
−
+
⋅
−
−
=
(
)
3
2
2
3
2
2 3
2
2
4
x
x
x
−
+
+
This is defined for all real numbers
x > −32 Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
141
(ii)
Let y = log7 (log x) = log (log )
log7
x (by change of base formula) The function is defined for all real numbers x > 1 |
1 | 2444-2447 | Therefore
dy
dx =
1
1
1
1
2
2
2
2
1
1
(3
2)
(3
2)
(2
4)
(2
4)
2
2
d
d
x
x
x
x
dx
dx
−
− −
+
⋅
+
+ −
+
⋅
+
= 1
2 3
2
3
21
2
4
4
21
2
23
(
)
( )
(
)
x
x
x
+
⋅
−
+
⋅
−
−
=
(
)
3
2
2
3
2
2 3
2
2
4
x
x
x
−
+
+
This is defined for all real numbers
x > −32 Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
141
(ii)
Let y = log7 (log x) = log (log )
log7
x (by change of base formula) The function is defined for all real numbers x > 1 Therefore
dy
dx =
1
(log (log ))
log7
d
x
dx
=
1
1
(log )
log7 log
d
x
x dx
⋅
=
1
xlog7 log
x
Example 40 Differentiate the following w |
1 | 2445-2448 | Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
141
(ii)
Let y = log7 (log x) = log (log )
log7
x (by change of base formula) The function is defined for all real numbers x > 1 Therefore
dy
dx =
1
(log (log ))
log7
d
x
dx
=
1
1
(log )
log7 log
d
x
x dx
⋅
=
1
xlog7 log
x
Example 40 Differentiate the following w r |
1 | 2446-2449 | The function is defined for all real numbers x > 1 Therefore
dy
dx =
1
(log (log ))
log7
d
x
dx
=
1
1
(log )
log7 log
d
x
x dx
⋅
=
1
xlog7 log
x
Example 40 Differentiate the following w r t |
1 | 2447-2450 | Therefore
dy
dx =
1
(log (log ))
log7
d
x
dx
=
1
1
(log )
log7 log
d
x
x dx
⋅
=
1
xlog7 log
x
Example 40 Differentiate the following w r t x |
1 | 2448-2451 | r t x (i)
cos –1 (sin x)
(ii)
1
sin
tan
1
cos
x
x
−
+
(iii)
1
1
2
sin
1
4
x
x
+
−
+
Solution
(i)
Let f (x) = cos –1 (sin x) |
1 | 2449-2452 | t x (i)
cos –1 (sin x)
(ii)
1
sin
tan
1
cos
x
x
−
+
(iii)
1
1
2
sin
1
4
x
x
+
−
+
Solution
(i)
Let f (x) = cos –1 (sin x) Observe that this function is defined for all real numbers |
1 | 2450-2453 | x (i)
cos –1 (sin x)
(ii)
1
sin
tan
1
cos
x
x
−
+
(iii)
1
1
2
sin
1
4
x
x
+
−
+
Solution
(i)
Let f (x) = cos –1 (sin x) Observe that this function is defined for all real numbers We may rewrite this function as
f(x) = cos –1 (sin x)
= cos
cos
−
−
1
π2
x
= 2
x
π −
Thus
f ′(x) = – 1 |
1 | 2451-2454 | (i)
cos –1 (sin x)
(ii)
1
sin
tan
1
cos
x
x
−
+
(iii)
1
1
2
sin
1
4
x
x
+
−
+
Solution
(i)
Let f (x) = cos –1 (sin x) Observe that this function is defined for all real numbers We may rewrite this function as
f(x) = cos –1 (sin x)
= cos
cos
−
−
1
π2
x
= 2
x
π −
Thus
f ′(x) = – 1 (ii)
Let f(x) = tan –1
sin
1
cos
x
x
+
|
1 | 2452-2455 | Observe that this function is defined for all real numbers We may rewrite this function as
f(x) = cos –1 (sin x)
= cos
cos
−
−
1
π2
x
= 2
x
π −
Thus
f ′(x) = – 1 (ii)
Let f(x) = tan –1
sin
1
cos
x
x
+
Observe that this function is defined for all real
numbers, where cos x ≠ – 1; i |
1 | 2453-2456 | We may rewrite this function as
f(x) = cos –1 (sin x)
= cos
cos
−
−
1
π2
x
= 2
x
π −
Thus
f ′(x) = – 1 (ii)
Let f(x) = tan –1
sin
1
cos
x
x
+
Observe that this function is defined for all real
numbers, where cos x ≠ – 1; i e |
1 | 2454-2457 | (ii)
Let f(x) = tan –1
sin
1
cos
x
x
+
Observe that this function is defined for all real
numbers, where cos x ≠ – 1; i e , at all odd multiplies of π |
1 | 2455-2458 | Observe that this function is defined for all real
numbers, where cos x ≠ – 1; i e , at all odd multiplies of π We may rewrite this
function as
f(x) =
1
sin
tan
1
cos
x
x
−
+
=
1
2
2 sin
2cos
2
tan
2cos 2
x
x
x
−
Rationalised 2023-24
MATHEMATICS
142
=
tan1
tan 2
2
x
x
−
=
Observe that we could cancel cos
x2
in both numerator and denominator as it
is not equal to zero |
1 | 2456-2459 | e , at all odd multiplies of π We may rewrite this
function as
f(x) =
1
sin
tan
1
cos
x
x
−
+
=
1
2
2 sin
2cos
2
tan
2cos 2
x
x
x
−
Rationalised 2023-24
MATHEMATICS
142
=
tan1
tan 2
2
x
x
−
=
Observe that we could cancel cos
x2
in both numerator and denominator as it
is not equal to zero Thus f ′(x) = 1 |
1 | 2457-2460 | , at all odd multiplies of π We may rewrite this
function as
f(x) =
1
sin
tan
1
cos
x
x
−
+
=
1
2
2 sin
2cos
2
tan
2cos 2
x
x
x
−
Rationalised 2023-24
MATHEMATICS
142
=
tan1
tan 2
2
x
x
−
=
Observe that we could cancel cos
x2
in both numerator and denominator as it
is not equal to zero Thus f ′(x) = 1 2
(iii)
Let f(x) = sin–1
1
12
4
x
x
+
+
|
1 | 2458-2461 | We may rewrite this
function as
f(x) =
1
sin
tan
1
cos
x
x
−
+
=
1
2
2 sin
2cos
2
tan
2cos 2
x
x
x
−
Rationalised 2023-24
MATHEMATICS
142
=
tan1
tan 2
2
x
x
−
=
Observe that we could cancel cos
x2
in both numerator and denominator as it
is not equal to zero Thus f ′(x) = 1 2
(iii)
Let f(x) = sin–1
1
12
4
x
x
+
+
To find the domain of this function we need to find all
x such that
21
1
1
1
4
x
x
+
− ≤
≤
+ |
1 | 2459-2462 | Thus f ′(x) = 1 2
(iii)
Let f(x) = sin–1
1
12
4
x
x
+
+
To find the domain of this function we need to find all
x such that
21
1
1
1
4
x
x
+
− ≤
≤
+ Since the quantity in the middle is always positive,
we need to find all x such that
1
2
1
1
4
x
x
+
≤
+
, i |
1 | 2460-2463 | 2
(iii)
Let f(x) = sin–1
1
12
4
x
x
+
+
To find the domain of this function we need to find all
x such that
21
1
1
1
4
x
x
+
− ≤
≤
+ Since the quantity in the middle is always positive,
we need to find all x such that
1
2
1
1
4
x
x
+
≤
+
, i e |
1 | 2461-2464 | To find the domain of this function we need to find all
x such that
21
1
1
1
4
x
x
+
− ≤
≤
+ Since the quantity in the middle is always positive,
we need to find all x such that
1
2
1
1
4
x
x
+
≤
+
, i e , all x such that 2x + 1 ≤ 1 + 4x |
1 | 2462-2465 | Since the quantity in the middle is always positive,
we need to find all x such that
1
2
1
1
4
x
x
+
≤
+
, i e , all x such that 2x + 1 ≤ 1 + 4x We
may rewrite this as 2 ≤ 1
2x + 2x which is true for all x |
1 | 2463-2466 | e , all x such that 2x + 1 ≤ 1 + 4x We
may rewrite this as 2 ≤ 1
2x + 2x which is true for all x Hence the function
is defined at every real number |
1 | 2464-2467 | , all x such that 2x + 1 ≤ 1 + 4x We
may rewrite this as 2 ≤ 1
2x + 2x which is true for all x Hence the function
is defined at every real number By putting 2x = tan θ, this function may be
rewritten as
f(x) =
1
1
2
sin
1
4
x
x
+
−
+
= sin−
⋅
+ (
)
1
2
2
2
1
2
x
x
=
1
2tan2
sin
1
tan
−
θ
+
θ
= sin –1 [sin 2θ]
= 2θ = 2 tan – 1 (2x)
Thus
f ′(x) =
(
)
2
1
2
(2 )
1
2
x
x
dxd
⋅
⋅
+
=
2
(2 )log2
1
4
x
x ⋅
+
=
1
2
log2
1
4
x
x
+
+
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
143
Example 41 Find f ′(x) if f(x) = (sin x)sin x for all 0 < x < π |
1 | 2465-2468 | We
may rewrite this as 2 ≤ 1
2x + 2x which is true for all x Hence the function
is defined at every real number By putting 2x = tan θ, this function may be
rewritten as
f(x) =
1
1
2
sin
1
4
x
x
+
−
+
= sin−
⋅
+ (
)
1
2
2
2
1
2
x
x
=
1
2tan2
sin
1
tan
−
θ
+
θ
= sin –1 [sin 2θ]
= 2θ = 2 tan – 1 (2x)
Thus
f ′(x) =
(
)
2
1
2
(2 )
1
2
x
x
dxd
⋅
⋅
+
=
2
(2 )log2
1
4
x
x ⋅
+
=
1
2
log2
1
4
x
x
+
+
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
143
Example 41 Find f ′(x) if f(x) = (sin x)sin x for all 0 < x < π Solution The function y = (sin x)sin x is defined for all positive real numbers |
1 | 2466-2469 | Hence the function
is defined at every real number By putting 2x = tan θ, this function may be
rewritten as
f(x) =
1
1
2
sin
1
4
x
x
+
−
+
= sin−
⋅
+ (
)
1
2
2
2
1
2
x
x
=
1
2tan2
sin
1
tan
−
θ
+
θ
= sin –1 [sin 2θ]
= 2θ = 2 tan – 1 (2x)
Thus
f ′(x) =
(
)
2
1
2
(2 )
1
2
x
x
dxd
⋅
⋅
+
=
2
(2 )log2
1
4
x
x ⋅
+
=
1
2
log2
1
4
x
x
+
+
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
143
Example 41 Find f ′(x) if f(x) = (sin x)sin x for all 0 < x < π Solution The function y = (sin x)sin x is defined for all positive real numbers Taking
logarithms, we have
log y = log (sin x)sin x = sin x log (sin x)
Then
1 dy
y dx = d
dx (sin x log (sin x))
= cos x log (sin x) + sin x |
1 | 2467-2470 | By putting 2x = tan θ, this function may be
rewritten as
f(x) =
1
1
2
sin
1
4
x
x
+
−
+
= sin−
⋅
+ (
)
1
2
2
2
1
2
x
x
=
1
2tan2
sin
1
tan
−
θ
+
θ
= sin –1 [sin 2θ]
= 2θ = 2 tan – 1 (2x)
Thus
f ′(x) =
(
)
2
1
2
(2 )
1
2
x
x
dxd
⋅
⋅
+
=
2
(2 )log2
1
4
x
x ⋅
+
=
1
2
log2
1
4
x
x
+
+
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
143
Example 41 Find f ′(x) if f(x) = (sin x)sin x for all 0 < x < π Solution The function y = (sin x)sin x is defined for all positive real numbers Taking
logarithms, we have
log y = log (sin x)sin x = sin x log (sin x)
Then
1 dy
y dx = d
dx (sin x log (sin x))
= cos x log (sin x) + sin x 1
(sin )
sin
d
x
x dx
⋅
= cos x log (sin x) + cos x
= (1 + log (sin x)) cos x
Thus
dy
dx = y((1 + log (sin x)) cos x) = (1 + log (sin x)) ( sin x)sin x cos x
Example 42 For a positive constant a find dy
dx , where
1
1
, and
a
t t
y
a
x
t
t
+
=
=
+
Solution Observe that both y and x are defined for all real t ≠ 0 |
1 | 2468-2471 | Solution The function y = (sin x)sin x is defined for all positive real numbers Taking
logarithms, we have
log y = log (sin x)sin x = sin x log (sin x)
Then
1 dy
y dx = d
dx (sin x log (sin x))
= cos x log (sin x) + sin x 1
(sin )
sin
d
x
x dx
⋅
= cos x log (sin x) + cos x
= (1 + log (sin x)) cos x
Thus
dy
dx = y((1 + log (sin x)) cos x) = (1 + log (sin x)) ( sin x)sin x cos x
Example 42 For a positive constant a find dy
dx , where
1
1
, and
a
t t
y
a
x
t
t
+
=
=
+
Solution Observe that both y and x are defined for all real t ≠ 0 Clearly
dy
dt =
(
)
t t1
d
a
dt
+
=
1
1
log
at t d
t
a
dt
t
+
+
⋅
=
1
12
1
log
at t
a
t
+
−
Similarly
dx
dt =
1
1
1
a
d
a t
t
t
dt
t
−
+
⋅
+
=
1
2
1
1
1
a
a t
t
t
−
+
⋅
−
dx
dt ≠ 0 only if t ≠ ± 1 |
1 | 2469-2472 | Taking
logarithms, we have
log y = log (sin x)sin x = sin x log (sin x)
Then
1 dy
y dx = d
dx (sin x log (sin x))
= cos x log (sin x) + sin x 1
(sin )
sin
d
x
x dx
⋅
= cos x log (sin x) + cos x
= (1 + log (sin x)) cos x
Thus
dy
dx = y((1 + log (sin x)) cos x) = (1 + log (sin x)) ( sin x)sin x cos x
Example 42 For a positive constant a find dy
dx , where
1
1
, and
a
t t
y
a
x
t
t
+
=
=
+
Solution Observe that both y and x are defined for all real t ≠ 0 Clearly
dy
dt =
(
)
t t1
d
a
dt
+
=
1
1
log
at t d
t
a
dt
t
+
+
⋅
=
1
12
1
log
at t
a
t
+
−
Similarly
dx
dt =
1
1
1
a
d
a t
t
t
dt
t
−
+
⋅
+
=
1
2
1
1
1
a
a t
t
t
−
+
⋅
−
dx
dt ≠ 0 only if t ≠ ± 1 Thus for t ≠ ± 1,
Rationalised 2023-24
MATHEMATICS
144
dy
dy
dt
dx
dx
dt
=
=
a
t
a
a t
t
t
t t
a
+
−
−
+
⋅
−
1
2
1
2
1
1
1
1
1
log
=
1
1
log
1
t t
a
a
a
a t
t
+
−
+
Example 43 Differentiate sin2 x w |
1 | 2470-2473 | 1
(sin )
sin
d
x
x dx
⋅
= cos x log (sin x) + cos x
= (1 + log (sin x)) cos x
Thus
dy
dx = y((1 + log (sin x)) cos x) = (1 + log (sin x)) ( sin x)sin x cos x
Example 42 For a positive constant a find dy
dx , where
1
1
, and
a
t t
y
a
x
t
t
+
=
=
+
Solution Observe that both y and x are defined for all real t ≠ 0 Clearly
dy
dt =
(
)
t t1
d
a
dt
+
=
1
1
log
at t d
t
a
dt
t
+
+
⋅
=
1
12
1
log
at t
a
t
+
−
Similarly
dx
dt =
1
1
1
a
d
a t
t
t
dt
t
−
+
⋅
+
=
1
2
1
1
1
a
a t
t
t
−
+
⋅
−
dx
dt ≠ 0 only if t ≠ ± 1 Thus for t ≠ ± 1,
Rationalised 2023-24
MATHEMATICS
144
dy
dy
dt
dx
dx
dt
=
=
a
t
a
a t
t
t
t t
a
+
−
−
+
⋅
−
1
2
1
2
1
1
1
1
1
log
=
1
1
log
1
t t
a
a
a
a t
t
+
−
+
Example 43 Differentiate sin2 x w r |
1 | 2471-2474 | Clearly
dy
dt =
(
)
t t1
d
a
dt
+
=
1
1
log
at t d
t
a
dt
t
+
+
⋅
=
1
12
1
log
at t
a
t
+
−
Similarly
dx
dt =
1
1
1
a
d
a t
t
t
dt
t
−
+
⋅
+
=
1
2
1
1
1
a
a t
t
t
−
+
⋅
−
dx
dt ≠ 0 only if t ≠ ± 1 Thus for t ≠ ± 1,
Rationalised 2023-24
MATHEMATICS
144
dy
dy
dt
dx
dx
dt
=
=
a
t
a
a t
t
t
t t
a
+
−
−
+
⋅
−
1
2
1
2
1
1
1
1
1
log
=
1
1
log
1
t t
a
a
a
a t
t
+
−
+
Example 43 Differentiate sin2 x w r t |
1 | 2472-2475 | Thus for t ≠ ± 1,
Rationalised 2023-24
MATHEMATICS
144
dy
dy
dt
dx
dx
dt
=
=
a
t
a
a t
t
t
t t
a
+
−
−
+
⋅
−
1
2
1
2
1
1
1
1
1
log
=
1
1
log
1
t t
a
a
a
a t
t
+
−
+
Example 43 Differentiate sin2 x w r t e cos x |
1 | 2473-2476 | r t e cos x Solution Let u (x) = sin2 x and v (x) = e cos x |
1 | 2474-2477 | t e cos x Solution Let u (x) = sin2 x and v (x) = e cos x We want to find
/
/
du
du dx
dv
=dv dx |
1 | 2475-2478 | e cos x Solution Let u (x) = sin2 x and v (x) = e cos x We want to find
/
/
du
du dx
dv
=dv dx Clearly
du
dx = 2 sin x cos x and dv
dx = e cos x (– sin x) = – (sin x) e cos x
Thus
du
dv =
cos
cos
2sin
cos
2cos
sin
x
x
x
x
x
x e
= −e
−
Miscellaneous Exercise on Chapter 5
Differentiate w |
1 | 2476-2479 | Solution Let u (x) = sin2 x and v (x) = e cos x We want to find
/
/
du
du dx
dv
=dv dx Clearly
du
dx = 2 sin x cos x and dv
dx = e cos x (– sin x) = – (sin x) e cos x
Thus
du
dv =
cos
cos
2sin
cos
2cos
sin
x
x
x
x
x
x e
= −e
−
Miscellaneous Exercise on Chapter 5
Differentiate w r |
1 | 2477-2480 | We want to find
/
/
du
du dx
dv
=dv dx Clearly
du
dx = 2 sin x cos x and dv
dx = e cos x (– sin x) = – (sin x) e cos x
Thus
du
dv =
cos
cos
2sin
cos
2cos
sin
x
x
x
x
x
x e
= −e
−
Miscellaneous Exercise on Chapter 5
Differentiate w r t |
1 | 2478-2481 | Clearly
du
dx = 2 sin x cos x and dv
dx = e cos x (– sin x) = – (sin x) e cos x
Thus
du
dv =
cos
cos
2sin
cos
2cos
sin
x
x
x
x
x
x e
= −e
−
Miscellaneous Exercise on Chapter 5
Differentiate w r t x the function in Exercises 1 to 11 |
1 | 2479-2482 | r t x the function in Exercises 1 to 11 1 |
1 | 2480-2483 | t x the function in Exercises 1 to 11 1 (3x2 – 9x + 5)9
2 |
1 | 2481-2484 | x the function in Exercises 1 to 11 1 (3x2 – 9x + 5)9
2 sin3 x + cos6 x
3 |
1 | 2482-2485 | 1 (3x2 – 9x + 5)9
2 sin3 x + cos6 x
3 (5x)3 cos 2x
4 |
1 | 2483-2486 | (3x2 – 9x + 5)9
2 sin3 x + cos6 x
3 (5x)3 cos 2x
4 sin–1(x
x ), 0 ≤ x ≤ 1
5 |
1 | 2484-2487 | sin3 x + cos6 x
3 (5x)3 cos 2x
4 sin–1(x
x ), 0 ≤ x ≤ 1
5 cos1
2
2
7
x
x
−
+
, – 2 < x < 2
6 |
1 | 2485-2488 | (5x)3 cos 2x
4 sin–1(x
x ), 0 ≤ x ≤ 1
5 cos1
2
2
7
x
x
−
+
, – 2 < x < 2
6 1
1
sin
1 sin
cot
1
sin
1 sin
x
x
x
x
−
+
+
−
+
−
−
, 0 < x < 2
π
7 |
1 | 2486-2489 | sin–1(x
x ), 0 ≤ x ≤ 1
5 cos1
2
2
7
x
x
−
+
, – 2 < x < 2
6 1
1
sin
1 sin
cot
1
sin
1 sin
x
x
x
x
−
+
+
−
+
−
−
, 0 < x < 2
π
7 (log x)log x, x > 1
8 |
1 | 2487-2490 | cos1
2
2
7
x
x
−
+
, – 2 < x < 2
6 1
1
sin
1 sin
cot
1
sin
1 sin
x
x
x
x
−
+
+
−
+
−
−
, 0 < x < 2
π
7 (log x)log x, x > 1
8 cos (a cos x + b sin x), for some constant a and b |
1 | 2488-2491 | 1
1
sin
1 sin
cot
1
sin
1 sin
x
x
x
x
−
+
+
−
+
−
−
, 0 < x < 2
π
7 (log x)log x, x > 1
8 cos (a cos x + b sin x), for some constant a and b 9 |
1 | 2489-2492 | (log x)log x, x > 1
8 cos (a cos x + b sin x), for some constant a and b 9 (sin x – cos x) (sin x – cos x),
3
4
4
x
π
π
<
<
10 |
1 | 2490-2493 | cos (a cos x + b sin x), for some constant a and b 9 (sin x – cos x) (sin x – cos x),
3
4
4
x
π
π
<
<
10 xx + xa + ax + aa, for some fixed a > 0 and x > 0
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
145
11 |
1 | 2491-2494 | 9 (sin x – cos x) (sin x – cos x),
3
4
4
x
π
π
<
<
10 xx + xa + ax + aa, for some fixed a > 0 and x > 0
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
145
11 (
)
2
2 3
3
x
xx
− +x
−
, for x > 3
12 |
1 | 2492-2495 | (sin x – cos x) (sin x – cos x),
3
4
4
x
π
π
<
<
10 xx + xa + ax + aa, for some fixed a > 0 and x > 0
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
145
11 (
)
2
2 3
3
x
xx
− +x
−
, for x > 3
12 Find dy
dx
, if y = 12 (1 – cos t), x = 10 (t – sin t),
2
2
t
π
π
−
< <
13 |
1 | 2493-2496 | xx + xa + ax + aa, for some fixed a > 0 and x > 0
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
145
11 (
)
2
2 3
3
x
xx
− +x
−
, for x > 3
12 Find dy
dx
, if y = 12 (1 – cos t), x = 10 (t – sin t),
2
2
t
π
π
−
< <
13 Find dy
dx , if y = sin–1 x + sin–1
2
1
−x
, 0 < x < 1
14 |
1 | 2494-2497 | (
)
2
2 3
3
x
xx
− +x
−
, for x > 3
12 Find dy
dx
, if y = 12 (1 – cos t), x = 10 (t – sin t),
2
2
t
π
π
−
< <
13 Find dy
dx , if y = sin–1 x + sin–1
2
1
−x
, 0 < x < 1
14 If
1
1
0
x
y
y
x
+
+
+
=
, for , – 1 < x < 1, prove that
(
)2
1
1
dy
dx
x
= −
+
15 |
1 | 2495-2498 | Find dy
dx
, if y = 12 (1 – cos t), x = 10 (t – sin t),
2
2
t
π
π
−
< <
13 Find dy
dx , if y = sin–1 x + sin–1
2
1
−x
, 0 < x < 1
14 If
1
1
0
x
y
y
x
+
+
+
=
, for , – 1 < x < 1, prove that
(
)2
1
1
dy
dx
x
= −
+
15 If (x – a)2 + (y – b)2 = c2, for some c > 0, prove that
3
2
2
2
2
1
dy
dx
d y
dx
+
is a constant independent of a and b |
1 | 2496-2499 | Find dy
dx , if y = sin–1 x + sin–1
2
1
−x
, 0 < x < 1
14 If
1
1
0
x
y
y
x
+
+
+
=
, for , – 1 < x < 1, prove that
(
)2
1
1
dy
dx
x
= −
+
15 If (x – a)2 + (y – b)2 = c2, for some c > 0, prove that
3
2
2
2
2
1
dy
dx
d y
dx
+
is a constant independent of a and b 16 |
1 | 2497-2500 | If
1
1
0
x
y
y
x
+
+
+
=
, for , – 1 < x < 1, prove that
(
)2
1
1
dy
dx
x
= −
+
15 If (x – a)2 + (y – b)2 = c2, for some c > 0, prove that
3
2
2
2
2
1
dy
dx
d y
dx
+
is a constant independent of a and b 16 If cos y = x cos (a + y), with cos a ≠ ± 1, prove that
cos (2
)
sin
dy
a
y
dx
+a
= |
1 | 2498-2501 | If (x – a)2 + (y – b)2 = c2, for some c > 0, prove that
3
2
2
2
2
1
dy
dx
d y
dx
+
is a constant independent of a and b 16 If cos y = x cos (a + y), with cos a ≠ ± 1, prove that
cos (2
)
sin
dy
a
y
dx
+a
= 17 |
1 | 2499-2502 | 16 If cos y = x cos (a + y), with cos a ≠ ± 1, prove that
cos (2
)
sin
dy
a
y
dx
+a
= 17 If x = a (cos t + t sin t) and y = a (sin t – t cos t), find
2
2
d y
dx |
1 | 2500-2503 | If cos y = x cos (a + y), with cos a ≠ ± 1, prove that
cos (2
)
sin
dy
a
y
dx
+a
= 17 If x = a (cos t + t sin t) and y = a (sin t – t cos t), find
2
2
d y
dx 18 |
1 | 2501-2504 | 17 If x = a (cos t + t sin t) and y = a (sin t – t cos t), find
2
2
d y
dx 18 If f(x) = | x |3, show that f ″(x) exists for all real x and find it |
1 | 2502-2505 | If x = a (cos t + t sin t) and y = a (sin t – t cos t), find
2
2
d y
dx 18 If f(x) = | x |3, show that f ″(x) exists for all real x and find it 19 |
1 | 2503-2506 | 18 If f(x) = | x |3, show that f ″(x) exists for all real x and find it 19 Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation,
obtain the sum formula for cosines |
1 | 2504-2507 | If f(x) = | x |3, show that f ″(x) exists for all real x and find it 19 Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation,
obtain the sum formula for cosines 20 |
1 | 2505-2508 | 19 Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation,
obtain the sum formula for cosines 20 Does there exist a function which is continuous everywhere but not differentiable
at exactly two points |
1 | 2506-2509 | Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation,
obtain the sum formula for cosines 20 Does there exist a function which is continuous everywhere but not differentiable
at exactly two points Justify your answer |
1 | 2507-2510 | 20 Does there exist a function which is continuous everywhere but not differentiable
at exactly two points Justify your answer 21 |
1 | 2508-2511 | Does there exist a function which is continuous everywhere but not differentiable
at exactly two points Justify your answer 21 If
( )
( )
( )
f x
g x
h x
y
l
m
n
a
b
c
=
, prove that
( )
( )
( )
f
x
g x
h x
dy
l
m
n
dx
a
b
c
′
′
′
=
22 |
1 | 2509-2512 | Justify your answer 21 If
( )
( )
( )
f x
g x
h x
y
l
m
n
a
b
c
=
, prove that
( )
( )
( )
f
x
g x
h x
dy
l
m
n
dx
a
b
c
′
′
′
=
22 If y =
acos1
x
e
−
, – 1 ≤ x ≤ 1, show that (
)
2
2
2
2
1
0
d y
dy
x
x
a y
dx
dx
−
−
−
= |
1 | 2510-2513 | 21 If
( )
( )
( )
f x
g x
h x
y
l
m
n
a
b
c
=
, prove that
( )
( )
( )
f
x
g x
h x
dy
l
m
n
dx
a
b
c
′
′
′
=
22 If y =
acos1
x
e
−
, – 1 ≤ x ≤ 1, show that (
)
2
2
2
2
1
0
d y
dy
x
x
a y
dx
dx
−
−
−
= Rationalised 2023-24
MATHEMATICS
146
Summary
® A real valued function is continuous at a point in its domain if the limit of the
function at that point equals the value of the function at that point |
1 | 2511-2514 | If
( )
( )
( )
f x
g x
h x
y
l
m
n
a
b
c
=
, prove that
( )
( )
( )
f
x
g x
h x
dy
l
m
n
dx
a
b
c
′
′
′
=
22 If y =
acos1
x
e
−
, – 1 ≤ x ≤ 1, show that (
)
2
2
2
2
1
0
d y
dy
x
x
a y
dx
dx
−
−
−
= Rationalised 2023-24
MATHEMATICS
146
Summary
® A real valued function is continuous at a point in its domain if the limit of the
function at that point equals the value of the function at that point A function
is continuous if it is continuous on the whole of its domain |
1 | 2512-2515 | If y =
acos1
x
e
−
, – 1 ≤ x ≤ 1, show that (
)
2
2
2
2
1
0
d y
dy
x
x
a y
dx
dx
−
−
−
= Rationalised 2023-24
MATHEMATICS
146
Summary
® A real valued function is continuous at a point in its domain if the limit of the
function at that point equals the value of the function at that point A function
is continuous if it is continuous on the whole of its domain ® Sum, difference, product and quotient of continuous functions are continuous |
1 | 2513-2516 | Rationalised 2023-24
MATHEMATICS
146
Summary
® A real valued function is continuous at a point in its domain if the limit of the
function at that point equals the value of the function at that point A function
is continuous if it is continuous on the whole of its domain ® Sum, difference, product and quotient of continuous functions are continuous i |
1 | 2514-2517 | A function
is continuous if it is continuous on the whole of its domain ® Sum, difference, product and quotient of continuous functions are continuous i e |
1 | 2515-2518 | ® Sum, difference, product and quotient of continuous functions are continuous i e , if f and g are continuous functions, then
(f ± g) (x) = f (x) ± g(x) is continuous |
1 | 2516-2519 | i e , if f and g are continuous functions, then
(f ± g) (x) = f (x) ± g(x) is continuous (f |
1 | 2517-2520 | e , if f and g are continuous functions, then
(f ± g) (x) = f (x) ± g(x) is continuous (f g) (x) = f (x) |
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