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1
|
2318-2321
|
Differentiating both sides w r t x, we have
1 dw
w dx
⋅
=
(log )
log
( )
d
d
x
x
x
x
dx
dx
+
⋅
=
1
log
1
x
x
⋅x
+
⋅
i
|
1
|
2319-2322
|
r t x, we have
1 dw
w dx
⋅
=
(log )
log
( )
d
d
x
x
x
x
dx
dx
+
⋅
=
1
log
1
x
x
⋅x
+
⋅
i e
|
1
|
2320-2323
|
t x, we have
1 dw
w dx
⋅
=
(log )
log
( )
d
d
x
x
x
x
dx
dx
+
⋅
=
1
log
1
x
x
⋅x
+
⋅
i e dw
dx = w (1 + log x)
= xx (1 + log x)
|
1
|
2321-2324
|
x, we have
1 dw
w dx
⋅
=
(log )
log
( )
d
d
x
x
x
x
dx
dx
+
⋅
=
1
log
1
x
x
⋅x
+
⋅
i e dw
dx = w (1 + log x)
= xx (1 + log x) (4)
From (1), (2), (3), (4), we have
log
log
x
y
x dy
y
dy
y
y
x
x
y dx
x
dx
+
+
+
+ xx (1 + log x) = 0
or
(x
|
1
|
2322-2325
|
e dw
dx = w (1 + log x)
= xx (1 + log x) (4)
From (1), (2), (3), (4), we have
log
log
x
y
x dy
y
dy
y
y
x
x
y dx
x
dx
+
+
+
+ xx (1 + log x) = 0
or
(x yx – 1 + xy
|
1
|
2323-2326
|
dw
dx = w (1 + log x)
= xx (1 + log x) (4)
From (1), (2), (3), (4), we have
log
log
x
y
x dy
y
dy
y
y
x
x
y dx
x
dx
+
+
+
+ xx (1 + log x) = 0
or
(x yx – 1 + xy log x) dy
dx = – xx (1 + log x) – y
|
1
|
2324-2327
|
(4)
From (1), (2), (3), (4), we have
log
log
x
y
x dy
y
dy
y
y
x
x
y dx
x
dx
+
+
+
+ xx (1 + log x) = 0
or
(x yx – 1 + xy log x) dy
dx = – xx (1 + log x) – y xy–1 – yx log y
Therefore
dy
dx =
1
1
[
log
|
1
|
2325-2328
|
yx – 1 + xy log x) dy
dx = – xx (1 + log x) – y xy–1 – yx log y
Therefore
dy
dx =
1
1
[
log (1
log )]
|
1
|
2326-2329
|
log x) dy
dx = – xx (1 + log x) – y xy–1 – yx log y
Therefore
dy
dx =
1
1
[
log (1
log )] log
x
y
x
x
y
y
y
y x
x
x
x y
x
x
−
−
−
+
+
+
+
Rationalised 2023-24
MATHEMATICS
134
EXERCISE 5
|
1
|
2327-2330
|
xy–1 – yx log y
Therefore
dy
dx =
1
1
[
log (1
log )] log
x
y
x
x
y
y
y
y x
x
x
x y
x
x
−
−
−
+
+
+
+
Rationalised 2023-24
MATHEMATICS
134
EXERCISE 5 5
Differentiate the functions given in Exercises 1 to 11 w
|
1
|
2328-2331
|
(1
log )] log
x
y
x
x
y
y
y
y x
x
x
x y
x
x
−
−
−
+
+
+
+
Rationalised 2023-24
MATHEMATICS
134
EXERCISE 5 5
Differentiate the functions given in Exercises 1 to 11 w r
|
1
|
2329-2332
|
log
x
y
x
x
y
y
y
y x
x
x
x y
x
x
−
−
−
+
+
+
+
Rationalised 2023-24
MATHEMATICS
134
EXERCISE 5 5
Differentiate the functions given in Exercises 1 to 11 w r t
|
1
|
2330-2333
|
5
Differentiate the functions given in Exercises 1 to 11 w r t x
|
1
|
2331-2334
|
r t x 1
|
1
|
2332-2335
|
t x 1 cos x
|
1
|
2333-2336
|
x 1 cos x cos 2x
|
1
|
2334-2337
|
1 cos x cos 2x cos 3x
2
|
1
|
2335-2338
|
cos x cos 2x cos 3x
2 (
1) (
2)
(
3) (
4) (
5)
x
x
x
x
x
−
−
−
−
−
3
|
1
|
2336-2339
|
cos 2x cos 3x
2 (
1) (
2)
(
3) (
4) (
5)
x
x
x
x
x
−
−
−
−
−
3 (log x)cos x
4
|
1
|
2337-2340
|
cos 3x
2 (
1) (
2)
(
3) (
4) (
5)
x
x
x
x
x
−
−
−
−
−
3 (log x)cos x
4 xx – 2sin x
5
|
1
|
2338-2341
|
(
1) (
2)
(
3) (
4) (
5)
x
x
x
x
x
−
−
−
−
−
3 (log x)cos x
4 xx – 2sin x
5 (x + 3)2
|
1
|
2339-2342
|
(log x)cos x
4 xx – 2sin x
5 (x + 3)2 (x + 4)3
|
1
|
2340-2343
|
xx – 2sin x
5 (x + 3)2 (x + 4)3 (x + 5)4
6
|
1
|
2341-2344
|
(x + 3)2 (x + 4)3 (x + 5)4
6 11
1
x
x
x
x
x
+
+
+
7
|
1
|
2342-2345
|
(x + 4)3 (x + 5)4
6 11
1
x
x
x
x
x
+
+
+
7 (log x)x + xlog x
8
|
1
|
2343-2346
|
(x + 5)4
6 11
1
x
x
x
x
x
+
+
+
7 (log x)x + xlog x
8 (sin x)x + sin–1
x
9
|
1
|
2344-2347
|
11
1
x
x
x
x
x
+
+
+
7 (log x)x + xlog x
8 (sin x)x + sin–1
x
9 xsin x + (sin x)cos x
10
|
1
|
2345-2348
|
(log x)x + xlog x
8 (sin x)x + sin–1
x
9 xsin x + (sin x)cos x
10 2
cos
2
1
1
x
x
x
x
x
+
+
−
11
|
1
|
2346-2349
|
(sin x)x + sin–1
x
9 xsin x + (sin x)cos x
10 2
cos
2
1
1
x
x
x
x
x
+
+
−
11 (x cos x)x +
1
( sin ) x
x
x
Find dy
dx of the functions given in Exercises 12 to 15
|
1
|
2347-2350
|
xsin x + (sin x)cos x
10 2
cos
2
1
1
x
x
x
x
x
+
+
−
11 (x cos x)x +
1
( sin ) x
x
x
Find dy
dx of the functions given in Exercises 12 to 15 12
|
1
|
2348-2351
|
2
cos
2
1
1
x
x
x
x
x
+
+
−
11 (x cos x)x +
1
( sin ) x
x
x
Find dy
dx of the functions given in Exercises 12 to 15 12 xy + yx = 1
13
|
1
|
2349-2352
|
(x cos x)x +
1
( sin ) x
x
x
Find dy
dx of the functions given in Exercises 12 to 15 12 xy + yx = 1
13 yx = xy
14
|
1
|
2350-2353
|
12 xy + yx = 1
13 yx = xy
14 (cos x)y = (cos y)x
15
|
1
|
2351-2354
|
xy + yx = 1
13 yx = xy
14 (cos x)y = (cos y)x
15 xy = e(x – y)
16
|
1
|
2352-2355
|
yx = xy
14 (cos x)y = (cos y)x
15 xy = e(x – y)
16 Find the derivative of the function given by f(x) = (1 + x) (1 + x2) (1 + x4) (1 + x8)
and hence find f′(1)
|
1
|
2353-2356
|
(cos x)y = (cos y)x
15 xy = e(x – y)
16 Find the derivative of the function given by f(x) = (1 + x) (1 + x2) (1 + x4) (1 + x8)
and hence find f′(1) 17
|
1
|
2354-2357
|
xy = e(x – y)
16 Find the derivative of the function given by f(x) = (1 + x) (1 + x2) (1 + x4) (1 + x8)
and hence find f′(1) 17 Differentiate (x2 – 5x + 8) (x3 + 7x + 9) in three ways mentioned below:
(i) by using product rule
(ii) by expanding the product to obtain a single polynomial
|
1
|
2355-2358
|
Find the derivative of the function given by f(x) = (1 + x) (1 + x2) (1 + x4) (1 + x8)
and hence find f′(1) 17 Differentiate (x2 – 5x + 8) (x3 + 7x + 9) in three ways mentioned below:
(i) by using product rule
(ii) by expanding the product to obtain a single polynomial (iii) by logarithmic differentiation
|
1
|
2356-2359
|
17 Differentiate (x2 – 5x + 8) (x3 + 7x + 9) in three ways mentioned below:
(i) by using product rule
(ii) by expanding the product to obtain a single polynomial (iii) by logarithmic differentiation Do they all give the same answer
|
1
|
2357-2360
|
Differentiate (x2 – 5x + 8) (x3 + 7x + 9) in three ways mentioned below:
(i) by using product rule
(ii) by expanding the product to obtain a single polynomial (iii) by logarithmic differentiation Do they all give the same answer 18
|
1
|
2358-2361
|
(iii) by logarithmic differentiation Do they all give the same answer 18 If u, v and w are functions of x, then show that
d
dx (u
|
1
|
2359-2362
|
Do they all give the same answer 18 If u, v and w are functions of x, then show that
d
dx (u v
|
1
|
2360-2363
|
18 If u, v and w are functions of x, then show that
d
dx (u v w) = du
dx v
|
1
|
2361-2364
|
If u, v and w are functions of x, then show that
d
dx (u v w) = du
dx v w + u
|
1
|
2362-2365
|
v w) = du
dx v w + u dv
dx
|
1
|
2363-2366
|
w) = du
dx v w + u dv
dx w + u
|
1
|
2364-2367
|
w + u dv
dx w + u v dw
dx
in two ways - first by repeated application of product rule, second by logarithmic
differentiation
|
1
|
2365-2368
|
dv
dx w + u v dw
dx
in two ways - first by repeated application of product rule, second by logarithmic
differentiation 5
|
1
|
2366-2369
|
w + u v dw
dx
in two ways - first by repeated application of product rule, second by logarithmic
differentiation 5 6 Derivatives of Functions in Parametric Forms
Sometimes the relation between two variables is neither explicit nor implicit, but some
link of a third variable with each of the two variables, separately, establishes a relation
between the first two variables
|
1
|
2367-2370
|
v dw
dx
in two ways - first by repeated application of product rule, second by logarithmic
differentiation 5 6 Derivatives of Functions in Parametric Forms
Sometimes the relation between two variables is neither explicit nor implicit, but some
link of a third variable with each of the two variables, separately, establishes a relation
between the first two variables In such a situation, we say that the relation between
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
135
them is expressed via a third variable
|
1
|
2368-2371
|
5 6 Derivatives of Functions in Parametric Forms
Sometimes the relation between two variables is neither explicit nor implicit, but some
link of a third variable with each of the two variables, separately, establishes a relation
between the first two variables In such a situation, we say that the relation between
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
135
them is expressed via a third variable The third variable is called the parameter
|
1
|
2369-2372
|
6 Derivatives of Functions in Parametric Forms
Sometimes the relation between two variables is neither explicit nor implicit, but some
link of a third variable with each of the two variables, separately, establishes a relation
between the first two variables In such a situation, we say that the relation between
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
135
them is expressed via a third variable The third variable is called the parameter More
precisely, a relation expressed between two variables x and y in the form
x = f(t), y = g (t) is said to be parametric form with t as a parameter
|
1
|
2370-2373
|
In such a situation, we say that the relation between
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
135
them is expressed via a third variable The third variable is called the parameter More
precisely, a relation expressed between two variables x and y in the form
x = f(t), y = g (t) is said to be parametric form with t as a parameter In order to find derivative of function in such form, we have by chain rule
|
1
|
2371-2374
|
The third variable is called the parameter More
precisely, a relation expressed between two variables x and y in the form
x = f(t), y = g (t) is said to be parametric form with t as a parameter In order to find derivative of function in such form, we have by chain rule dy
dt = dy dx
dx dt
⋅
or
dy
dx =
whenever
0
dy
dx
dt
dx
dt
dt
≠
Thus
dy
dx =
( )
as
( ) and
( )
g t( )
dy
dx
g t
f t
f t
dt
dt
′
=
′
=
′
′
[provided f′(t) ≠ 0]
Example 31 Find dy
dx , if x = a cos θ, y = a sin θ
|
1
|
2372-2375
|
More
precisely, a relation expressed between two variables x and y in the form
x = f(t), y = g (t) is said to be parametric form with t as a parameter In order to find derivative of function in such form, we have by chain rule dy
dt = dy dx
dx dt
⋅
or
dy
dx =
whenever
0
dy
dx
dt
dx
dt
dt
≠
Thus
dy
dx =
( )
as
( ) and
( )
g t( )
dy
dx
g t
f t
f t
dt
dt
′
=
′
=
′
′
[provided f′(t) ≠ 0]
Example 31 Find dy
dx , if x = a cos θ, y = a sin θ Solution Given that
x = a cos θ, y = a sin θ
Therefore
dx
dθ = – a sin θ, dy
dθ = a cos θ
Hence
dy
dx =
cos
cot
sin
dy
a
d
dx
a
d
θ
θ =
= −
θ
−
θ
θ
Example 32 Find dy
dx
, if x = at2, y = 2at
|
1
|
2373-2376
|
In order to find derivative of function in such form, we have by chain rule dy
dt = dy dx
dx dt
⋅
or
dy
dx =
whenever
0
dy
dx
dt
dx
dt
dt
≠
Thus
dy
dx =
( )
as
( ) and
( )
g t( )
dy
dx
g t
f t
f t
dt
dt
′
=
′
=
′
′
[provided f′(t) ≠ 0]
Example 31 Find dy
dx , if x = a cos θ, y = a sin θ Solution Given that
x = a cos θ, y = a sin θ
Therefore
dx
dθ = – a sin θ, dy
dθ = a cos θ
Hence
dy
dx =
cos
cot
sin
dy
a
d
dx
a
d
θ
θ =
= −
θ
−
θ
θ
Example 32 Find dy
dx
, if x = at2, y = 2at Solution Given that x = at2, y = 2at
So
dx
dt = 2at and dy
dt = 2a
Therefore
dy
dx =
2
1
2
dy
a
dt
dx
at
t
dt
=
=
Rationalised 2023-24
MATHEMATICS
136
Example 33 Find dy
dx
, if x = a (θ + sin θ), y = a (1 – cos θ)
|
1
|
2374-2377
|
dy
dt = dy dx
dx dt
⋅
or
dy
dx =
whenever
0
dy
dx
dt
dx
dt
dt
≠
Thus
dy
dx =
( )
as
( ) and
( )
g t( )
dy
dx
g t
f t
f t
dt
dt
′
=
′
=
′
′
[provided f′(t) ≠ 0]
Example 31 Find dy
dx , if x = a cos θ, y = a sin θ Solution Given that
x = a cos θ, y = a sin θ
Therefore
dx
dθ = – a sin θ, dy
dθ = a cos θ
Hence
dy
dx =
cos
cot
sin
dy
a
d
dx
a
d
θ
θ =
= −
θ
−
θ
θ
Example 32 Find dy
dx
, if x = at2, y = 2at Solution Given that x = at2, y = 2at
So
dx
dt = 2at and dy
dt = 2a
Therefore
dy
dx =
2
1
2
dy
a
dt
dx
at
t
dt
=
=
Rationalised 2023-24
MATHEMATICS
136
Example 33 Find dy
dx
, if x = a (θ + sin θ), y = a (1 – cos θ) Solution We have dx
dθ = a(1 + cos θ), dy
dθ = a (sin θ)
Therefore
dy
dx =
sin
tan
(1
cos )
2
dy
a
d
dx
a
d
θ
θ
θ =
=
+
θ
θ
ANote It may be noted here that dy
dx is expressed in terms of parameter only
without directly involving the main variables x and y
|
1
|
2375-2378
|
Solution Given that
x = a cos θ, y = a sin θ
Therefore
dx
dθ = – a sin θ, dy
dθ = a cos θ
Hence
dy
dx =
cos
cot
sin
dy
a
d
dx
a
d
θ
θ =
= −
θ
−
θ
θ
Example 32 Find dy
dx
, if x = at2, y = 2at Solution Given that x = at2, y = 2at
So
dx
dt = 2at and dy
dt = 2a
Therefore
dy
dx =
2
1
2
dy
a
dt
dx
at
t
dt
=
=
Rationalised 2023-24
MATHEMATICS
136
Example 33 Find dy
dx
, if x = a (θ + sin θ), y = a (1 – cos θ) Solution We have dx
dθ = a(1 + cos θ), dy
dθ = a (sin θ)
Therefore
dy
dx =
sin
tan
(1
cos )
2
dy
a
d
dx
a
d
θ
θ
θ =
=
+
θ
θ
ANote It may be noted here that dy
dx is expressed in terms of parameter only
without directly involving the main variables x and y Example 34 Find
2
2
2
3
3
3
dy, if
x
y
a
dx
+
=
|
1
|
2376-2379
|
Solution Given that x = at2, y = 2at
So
dx
dt = 2at and dy
dt = 2a
Therefore
dy
dx =
2
1
2
dy
a
dt
dx
at
t
dt
=
=
Rationalised 2023-24
MATHEMATICS
136
Example 33 Find dy
dx
, if x = a (θ + sin θ), y = a (1 – cos θ) Solution We have dx
dθ = a(1 + cos θ), dy
dθ = a (sin θ)
Therefore
dy
dx =
sin
tan
(1
cos )
2
dy
a
d
dx
a
d
θ
θ
θ =
=
+
θ
θ
ANote It may be noted here that dy
dx is expressed in terms of parameter only
without directly involving the main variables x and y Example 34 Find
2
2
2
3
3
3
dy, if
x
y
a
dx
+
= Solution Let x = a cos3 θ, y = a sin3 θ
|
1
|
2377-2380
|
Solution We have dx
dθ = a(1 + cos θ), dy
dθ = a (sin θ)
Therefore
dy
dx =
sin
tan
(1
cos )
2
dy
a
d
dx
a
d
θ
θ
θ =
=
+
θ
θ
ANote It may be noted here that dy
dx is expressed in terms of parameter only
without directly involving the main variables x and y Example 34 Find
2
2
2
3
3
3
dy, if
x
y
a
dx
+
= Solution Let x = a cos3 θ, y = a sin3 θ Then
2
2
3
3
x
+y
=
2
2
3
3
3
3
( cos
)
( sin
)
a
a
θ
+
θ
=
2
2
2
2
3
3
(cos
(sin
)
a
a
θ +
θ =
Hence, x = a cos3θ, y = a sin3θ is parametric equation of
2
2
2
3
3
3
x
y
a
+
=
Now
dx
dθ = – 3a cos2 θ sin θ and dy
dθ = 3a sin2 θ cos θ
Therefore
dy
dx =
2
3
3 sin2
cos
tan
3 cos
sin
dy
a
y
d
dx
x
a
d
θ
θ
θ =
= −
θ = −
−
θ
θ
θ
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
137
EXERCISE 5
|
1
|
2378-2381
|
Example 34 Find
2
2
2
3
3
3
dy, if
x
y
a
dx
+
= Solution Let x = a cos3 θ, y = a sin3 θ Then
2
2
3
3
x
+y
=
2
2
3
3
3
3
( cos
)
( sin
)
a
a
θ
+
θ
=
2
2
2
2
3
3
(cos
(sin
)
a
a
θ +
θ =
Hence, x = a cos3θ, y = a sin3θ is parametric equation of
2
2
2
3
3
3
x
y
a
+
=
Now
dx
dθ = – 3a cos2 θ sin θ and dy
dθ = 3a sin2 θ cos θ
Therefore
dy
dx =
2
3
3 sin2
cos
tan
3 cos
sin
dy
a
y
d
dx
x
a
d
θ
θ
θ =
= −
θ = −
−
θ
θ
θ
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
137
EXERCISE 5 6
If x and y are connected parametrically by the equations given in Exercises 1 to 10,
without eliminating the parameter, Find dy
dx
|
1
|
2379-2382
|
Solution Let x = a cos3 θ, y = a sin3 θ Then
2
2
3
3
x
+y
=
2
2
3
3
3
3
( cos
)
( sin
)
a
a
θ
+
θ
=
2
2
2
2
3
3
(cos
(sin
)
a
a
θ +
θ =
Hence, x = a cos3θ, y = a sin3θ is parametric equation of
2
2
2
3
3
3
x
y
a
+
=
Now
dx
dθ = – 3a cos2 θ sin θ and dy
dθ = 3a sin2 θ cos θ
Therefore
dy
dx =
2
3
3 sin2
cos
tan
3 cos
sin
dy
a
y
d
dx
x
a
d
θ
θ
θ =
= −
θ = −
−
θ
θ
θ
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
137
EXERCISE 5 6
If x and y are connected parametrically by the equations given in Exercises 1 to 10,
without eliminating the parameter, Find dy
dx 1
|
1
|
2380-2383
|
Then
2
2
3
3
x
+y
=
2
2
3
3
3
3
( cos
)
( sin
)
a
a
θ
+
θ
=
2
2
2
2
3
3
(cos
(sin
)
a
a
θ +
θ =
Hence, x = a cos3θ, y = a sin3θ is parametric equation of
2
2
2
3
3
3
x
y
a
+
=
Now
dx
dθ = – 3a cos2 θ sin θ and dy
dθ = 3a sin2 θ cos θ
Therefore
dy
dx =
2
3
3 sin2
cos
tan
3 cos
sin
dy
a
y
d
dx
x
a
d
θ
θ
θ =
= −
θ = −
−
θ
θ
θ
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
137
EXERCISE 5 6
If x and y are connected parametrically by the equations given in Exercises 1 to 10,
without eliminating the parameter, Find dy
dx 1 x = 2at2, y = at4
2
|
1
|
2381-2384
|
6
If x and y are connected parametrically by the equations given in Exercises 1 to 10,
without eliminating the parameter, Find dy
dx 1 x = 2at2, y = at4
2 x = a cos θ, y = b cos θ
3
|
1
|
2382-2385
|
1 x = 2at2, y = at4
2 x = a cos θ, y = b cos θ
3 x = sin t, y = cos 2t
4
|
1
|
2383-2386
|
x = 2at2, y = at4
2 x = a cos θ, y = b cos θ
3 x = sin t, y = cos 2t
4 x = 4t, y = 4
t
5
|
1
|
2384-2387
|
x = a cos θ, y = b cos θ
3 x = sin t, y = cos 2t
4 x = 4t, y = 4
t
5 x = cos θ – cos 2θ, y = sin θ – sin 2θ
6
|
1
|
2385-2388
|
x = sin t, y = cos 2t
4 x = 4t, y = 4
t
5 x = cos θ – cos 2θ, y = sin θ – sin 2θ
6 x = a (θ – sin θ), y = a (1 + cos θ)
7
|
1
|
2386-2389
|
x = 4t, y = 4
t
5 x = cos θ – cos 2θ, y = sin θ – sin 2θ
6 x = a (θ – sin θ), y = a (1 + cos θ)
7 x =
sin3
cos2
t
t ,
cos3
cos2
t
y
t
=
8
|
1
|
2387-2390
|
x = cos θ – cos 2θ, y = sin θ – sin 2θ
6 x = a (θ – sin θ), y = a (1 + cos θ)
7 x =
sin3
cos2
t
t ,
cos3
cos2
t
y
t
=
8 cos
log tan 2
t
x
a
t
=
+
y = a sin t
9
|
1
|
2388-2391
|
x = a (θ – sin θ), y = a (1 + cos θ)
7 x =
sin3
cos2
t
t ,
cos3
cos2
t
y
t
=
8 cos
log tan 2
t
x
a
t
=
+
y = a sin t
9 x = a sec θ, y = b tan θ
10
|
1
|
2389-2392
|
x =
sin3
cos2
t
t ,
cos3
cos2
t
y
t
=
8 cos
log tan 2
t
x
a
t
=
+
y = a sin t
9 x = a sec θ, y = b tan θ
10 x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ)
11
|
1
|
2390-2393
|
cos
log tan 2
t
x
a
t
=
+
y = a sin t
9 x = a sec θ, y = b tan θ
10 x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ)
11 If
1
1
sin
cos
,
, show that
t
t
dy
y
x
a
y
a
dx
x
−
−
=
=
= −
5
|
1
|
2391-2394
|
x = a sec θ, y = b tan θ
10 x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ)
11 If
1
1
sin
cos
,
, show that
t
t
dy
y
x
a
y
a
dx
x
−
−
=
=
= −
5 7 Second Order Derivative
Let
y = f (x)
|
1
|
2392-2395
|
x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ)
11 If
1
1
sin
cos
,
, show that
t
t
dy
y
x
a
y
a
dx
x
−
−
=
=
= −
5 7 Second Order Derivative
Let
y = f (x) Then
dy
dx = f ′(x)
|
1
|
2393-2396
|
If
1
1
sin
cos
,
, show that
t
t
dy
y
x
a
y
a
dx
x
−
−
=
=
= −
5 7 Second Order Derivative
Let
y = f (x) Then
dy
dx = f ′(x) (1)
If f′(x) is differentiable, we may differentiate (1) again w
|
1
|
2394-2397
|
7 Second Order Derivative
Let
y = f (x) Then
dy
dx = f ′(x) (1)
If f′(x) is differentiable, we may differentiate (1) again w r
|
1
|
2395-2398
|
Then
dy
dx = f ′(x) (1)
If f′(x) is differentiable, we may differentiate (1) again w r t
|
1
|
2396-2399
|
(1)
If f′(x) is differentiable, we may differentiate (1) again w r t x
|
1
|
2397-2400
|
r t x Then, the left hand
side becomes d
dy
dx
dx
which is called the second order derivative of y w
|
1
|
2398-2401
|
t x Then, the left hand
side becomes d
dy
dx
dx
which is called the second order derivative of y w r
|
1
|
2399-2402
|
x Then, the left hand
side becomes d
dy
dx
dx
which is called the second order derivative of y w r t
|
1
|
2400-2403
|
Then, the left hand
side becomes d
dy
dx
dx
which is called the second order derivative of y w r t x and
is denoted by
2
2
d y
dx
|
1
|
2401-2404
|
r t x and
is denoted by
2
2
d y
dx The second order derivative of f(x) is denoted by f ″(x)
|
1
|
2402-2405
|
t x and
is denoted by
2
2
d y
dx The second order derivative of f(x) is denoted by f ″(x) It is also
denoted by D2 y or y″ or y2 if y = f(x)
|
1
|
2403-2406
|
x and
is denoted by
2
2
d y
dx The second order derivative of f(x) is denoted by f ″(x) It is also
denoted by D2 y or y″ or y2 if y = f(x) We remark that higher order derivatives may be
defined similarly
|
1
|
2404-2407
|
The second order derivative of f(x) is denoted by f ″(x) It is also
denoted by D2 y or y″ or y2 if y = f(x) We remark that higher order derivatives may be
defined similarly Rationalised 2023-24
MATHEMATICS
138
Example 35 Find
2
2
d y
dx
, if y = x3 + tan x
|
1
|
2405-2408
|
It is also
denoted by D2 y or y″ or y2 if y = f(x) We remark that higher order derivatives may be
defined similarly Rationalised 2023-24
MATHEMATICS
138
Example 35 Find
2
2
d y
dx
, if y = x3 + tan x Solution Given that y = x3 + tan x
|
1
|
2406-2409
|
We remark that higher order derivatives may be
defined similarly Rationalised 2023-24
MATHEMATICS
138
Example 35 Find
2
2
d y
dx
, if y = x3 + tan x Solution Given that y = x3 + tan x Then
dy
dx = 3x2 + sec2 x
Therefore
2
2
d y
dx
=
(
)
2
2
3
sec
d
x
x
dx
+
= 6x + 2 sec x
|
1
|
2407-2410
|
Rationalised 2023-24
MATHEMATICS
138
Example 35 Find
2
2
d y
dx
, if y = x3 + tan x Solution Given that y = x3 + tan x Then
dy
dx = 3x2 + sec2 x
Therefore
2
2
d y
dx
=
(
)
2
2
3
sec
d
x
x
dx
+
= 6x + 2 sec x sec x tan x = 6x + 2 sec2 x tan x
Example 36 If y = A sin x + B cos x, then prove that
2
2
0
d y
y
dx
+
=
|
1
|
2408-2411
|
Solution Given that y = x3 + tan x Then
dy
dx = 3x2 + sec2 x
Therefore
2
2
d y
dx
=
(
)
2
2
3
sec
d
x
x
dx
+
= 6x + 2 sec x sec x tan x = 6x + 2 sec2 x tan x
Example 36 If y = A sin x + B cos x, then prove that
2
2
0
d y
y
dx
+
= Solution We have
dy
dx = A cos x – B sin x
and
2
2
d y
dx
= d
dx (A cos x – B sin x)
= – A sin x – B cos x = – y
Hence
2
2
d y
dx
+ y = 0
Example 37 If y = 3e2x + 2e3x, prove that
2
2
5
6
0
d y
dy
y
dx
dx
−
+
=
|
1
|
2409-2412
|
Then
dy
dx = 3x2 + sec2 x
Therefore
2
2
d y
dx
=
(
)
2
2
3
sec
d
x
x
dx
+
= 6x + 2 sec x sec x tan x = 6x + 2 sec2 x tan x
Example 36 If y = A sin x + B cos x, then prove that
2
2
0
d y
y
dx
+
= Solution We have
dy
dx = A cos x – B sin x
and
2
2
d y
dx
= d
dx (A cos x – B sin x)
= – A sin x – B cos x = – y
Hence
2
2
d y
dx
+ y = 0
Example 37 If y = 3e2x + 2e3x, prove that
2
2
5
6
0
d y
dy
y
dx
dx
−
+
= Solution Given that y = 3e2x + 2e3x
|
1
|
2410-2413
|
sec x tan x = 6x + 2 sec2 x tan x
Example 36 If y = A sin x + B cos x, then prove that
2
2
0
d y
y
dx
+
= Solution We have
dy
dx = A cos x – B sin x
and
2
2
d y
dx
= d
dx (A cos x – B sin x)
= – A sin x – B cos x = – y
Hence
2
2
d y
dx
+ y = 0
Example 37 If y = 3e2x + 2e3x, prove that
2
2
5
6
0
d y
dy
y
dx
dx
−
+
= Solution Given that y = 3e2x + 2e3x Then
dy
dx = 6e2x + 6e3x = 6 (e2x + e3x)
Therefore
2
2
d y
dx
= 12e2x + 18e3x = 6 (2e2x + 3e3x)
Hence
2
2
5
d y
dy
dx
dx
−
+ 6y = 6 (2e2x + 3e3x)
– 30 (e2x + e3x) + 6 (3e2x + 2e3x) = 0
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
139
Example 38 If y = sin–1 x, show that (1 – x2)
2
2
0
d y
dy
x dx
dx
−
=
|
1
|
2411-2414
|
Solution We have
dy
dx = A cos x – B sin x
and
2
2
d y
dx
= d
dx (A cos x – B sin x)
= – A sin x – B cos x = – y
Hence
2
2
d y
dx
+ y = 0
Example 37 If y = 3e2x + 2e3x, prove that
2
2
5
6
0
d y
dy
y
dx
dx
−
+
= Solution Given that y = 3e2x + 2e3x Then
dy
dx = 6e2x + 6e3x = 6 (e2x + e3x)
Therefore
2
2
d y
dx
= 12e2x + 18e3x = 6 (2e2x + 3e3x)
Hence
2
2
5
d y
dy
dx
dx
−
+ 6y = 6 (2e2x + 3e3x)
– 30 (e2x + e3x) + 6 (3e2x + 2e3x) = 0
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
139
Example 38 If y = sin–1 x, show that (1 – x2)
2
2
0
d y
dy
x dx
dx
−
= Solution We have y = sin–1x
|
1
|
2412-2415
|
Solution Given that y = 3e2x + 2e3x Then
dy
dx = 6e2x + 6e3x = 6 (e2x + e3x)
Therefore
2
2
d y
dx
= 12e2x + 18e3x = 6 (2e2x + 3e3x)
Hence
2
2
5
d y
dy
dx
dx
−
+ 6y = 6 (2e2x + 3e3x)
– 30 (e2x + e3x) + 6 (3e2x + 2e3x) = 0
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
139
Example 38 If y = sin–1 x, show that (1 – x2)
2
2
0
d y
dy
x dx
dx
−
= Solution We have y = sin–1x Then
dy
dx =
2
1
(1
x)
−
or
2
(1
)
1
dy
x
dx
−
=
So
2
(1
)
|
1
|
2413-2416
|
Then
dy
dx = 6e2x + 6e3x = 6 (e2x + e3x)
Therefore
2
2
d y
dx
= 12e2x + 18e3x = 6 (2e2x + 3e3x)
Hence
2
2
5
d y
dy
dx
dx
−
+ 6y = 6 (2e2x + 3e3x)
– 30 (e2x + e3x) + 6 (3e2x + 2e3x) = 0
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
139
Example 38 If y = sin–1 x, show that (1 – x2)
2
2
0
d y
dy
x dx
dx
−
= Solution We have y = sin–1x Then
dy
dx =
2
1
(1
x)
−
or
2
(1
)
1
dy
x
dx
−
=
So
2
(1
) 0
d
dy
x
dx
dx
−
=
or
(
)
2
2
2
2
(1
)
(1
)
0
d y
dy
d
x
x
dx dx
dx
−
⋅
+
⋅
−
=
or
2
2
2
2
2
(1
)
0
2 1
d y
dy
x
x
dx
dx
x
−
⋅
−
⋅
=
−
Hence
2
2
2
(1
)
0
d y
dy
x
x dx
dx
−
−
=
Alternatively, Given that y = sin–1 x, we have
1
2
1
1
y
x
=
−
, i
|
1
|
2414-2417
|
Solution We have y = sin–1x Then
dy
dx =
2
1
(1
x)
−
or
2
(1
)
1
dy
x
dx
−
=
So
2
(1
) 0
d
dy
x
dx
dx
−
=
or
(
)
2
2
2
2
(1
)
(1
)
0
d y
dy
d
x
x
dx dx
dx
−
⋅
+
⋅
−
=
or
2
2
2
2
2
(1
)
0
2 1
d y
dy
x
x
dx
dx
x
−
⋅
−
⋅
=
−
Hence
2
2
2
(1
)
0
d y
dy
x
x dx
dx
−
−
=
Alternatively, Given that y = sin–1 x, we have
1
2
1
1
y
x
=
−
, i e
|
1
|
2415-2418
|
Then
dy
dx =
2
1
(1
x)
−
or
2
(1
)
1
dy
x
dx
−
=
So
2
(1
) 0
d
dy
x
dx
dx
−
=
or
(
)
2
2
2
2
(1
)
(1
)
0
d y
dy
d
x
x
dx dx
dx
−
⋅
+
⋅
−
=
or
2
2
2
2
2
(1
)
0
2 1
d y
dy
x
x
dx
dx
x
−
⋅
−
⋅
=
−
Hence
2
2
2
(1
)
0
d y
dy
x
x dx
dx
−
−
=
Alternatively, Given that y = sin–1 x, we have
1
2
1
1
y
x
=
−
, i e , (
2)
12
1
1
x
y
−
=
So
2
2
1
2
1
(1
)
|
1
|
2416-2419
|
0
d
dy
x
dx
dx
−
=
or
(
)
2
2
2
2
(1
)
(1
)
0
d y
dy
d
x
x
dx dx
dx
−
⋅
+
⋅
−
=
or
2
2
2
2
2
(1
)
0
2 1
d y
dy
x
x
dx
dx
x
−
⋅
−
⋅
=
−
Hence
2
2
2
(1
)
0
d y
dy
x
x dx
dx
−
−
=
Alternatively, Given that y = sin–1 x, we have
1
2
1
1
y
x
=
−
, i e , (
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
|
1
|
2417-2420
|
e , (
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
|
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