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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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