Loading [MathJax]/jax/output/CommonHTML/jax.js
MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 3 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Differentiation
Quiz 3
lf
y
=
log
(
log
(
x
+
√
1
+
x
2
)
)
then
d
y
d
x
=
Report Question
0%
1
x
+
√
1
+
x
2
0%
x
log
(
x
+
√
1
+
x
2
)
0%
−
1
log
(
x
+
√
1
+
x
2
)
0%
1
√
1
+
x
2
log
(
x
+
√
1
+
x
2
)
Explanation
We have,
y
=
log
(
log
(
x
+
√
1
+
x
2
)
)
Using chain rule of differentiation,
d
y
d
x
=
1
log
(
x
+
√
1
+
x
2
)
×
1
(
x
+
√
1
+
x
2
)
×
(
1
+
2
x
2
√
1
+
x
2
)
=
1
log
(
x
+
√
1
+
x
2
)
×
1
(
x
+
√
1
+
x
2
)
×
(
√
1
+
x
2
+
x
√
1
+
x
2
)
=
1
√
1
+
x
2
log
(
x
+
√
1
+
x
2
)
Hence,
d
y
d
x
=
1
√
1
+
x
2
log
(
x
+
√
1
+
x
2
)
lf
a
y
=
log
a
(
x
2
+
x
+
1
)
, then
d
y
d
x
=
Report Question
0%
log
a
e
.
(
2
x
+
1
)
(
x
2
+
x
+
1
)
log
(
x
2
+
x
+
1
)
0%
(
2
x
+
1
)
(
x
2
+
x
+
1
)
log
(
x
2
+
x
+
1
)
0%
1
(
x
2
+
x
+
1
)
log
(
x
2
+
x
+
1
)
0%
−
1
(
x
2
+
x
+
1
)
log
(
x
2
+
x
+
1
)
Explanation
We have,
a
y
=
log
a
(
x
2
+
x
+
1
)
⇒
(
1
)
Now differentiating both sides, w.r.t
x
⇒
a
y
(
log
a
)
d
y
d
x
=
log
a
e
⋅
2
x
+
1
x
2
+
x
+
1
⇒
d
y
d
x
=
log
a
e
⋅
(
2
x
+
1
)
(
x
2
+
x
+
1
)
⋅
log
(
x
2
+
x
+
1
)
, using (1)
If
√
x
+
y
+
√
y
−
x
=
c
, then
d
2
y
d
x
2
is
Report Question
0%
2
c
0%
−
2
c
2
0%
2
c
2
0%
−
2
c
Explanation
√
x
+
y
=
c
−
√
y
−
x
On squaring we get,
2
x
−
c
2
=
−
2
c
√
y
−
x
Again squaring we get,
y
=
(
2
x
−
c
2
)
2
4
c
2
+
x
y
′
=
4
(
2
x
−
c
2
)
4
c
2
+
1
y
″
=
8
4
c
2
=
2
c
2
Option C
If
f
(
x
)
=
(
a
x
+
b
)
cos
x
+
(
c
x
+
d
)
sin
x
and
f
′
(
x
)
=
x
cos
x
, for all values of
x
∈
R
, then
a
,
b
,
c
,
d
are given by
Report Question
0%
a
=
b
=
c
=
d
0%
0
,
1
,
−
1
,
0
0%
1
,
0
,
−
1
,
0
0%
0
,
1
,
1
,
0
Explanation
Given,
f
(
x
)
=
(
a
x
+
b
)
cos
x
+
(
c
x
+
d
)
sin
x
Thus
f
′
(
x
)
=
(
a
x
+
b
)
(
−
sin
x
)
+
a
cos
x
+
c
sin
x
+
(
c
x
+
d
)
cos
x
=
(
−
a
x
−
b
+
c
)
sin
x
+
(
a
+
c
x
+
d
)
cos
x
Also given,
f
′
(
x
)
=
x
cos
x
Comparing the coefficients, we get
a
=
0
,
b
=
1
,
c
=
1
,
d
=
0
The set onto which the derivative of the function
f
(
x
)
=
x
(
log
x
−
1
)
maps the ray
[
1
,
∞
)
is ?
Report Question
0%
[
1
,
∞
)
0%
(
10
,
∞
)
0%
[
0
,
∞
)
0%
(
0
,
0
)
Explanation
Given function
f
(
x
)
=
x
(
log
x
−
1
)
differentiate w.r.t
x
we get,
f
′
(
x
)
=
x
x
+
log
x
−
1
f
′
(
x
)
=
log
x
Clearly,
log
x
is defined in the given region.
as
x
→
1
,
log
x
=
0
x
→
∞
,
log
x
=
∞
The set onto which the derivative of the function maps is [
0
,
∞
).
lf
y
=
log
(
x
+
√
1
+
x
2
)
√
1
+
x
2
then
(
1
+
x
2
)
y
1
+
x
y
=
Report Question
0%
y
0%
−
y
0%
0
0%
1
Explanation
y
√
1
+
x
2
=
log
(
x
+
√
1
+
x
2
)
Differentiate both sides:
y
1
√
1
+
x
2
+
y
x
√
1
+
x
2
=
1
+
x
√
1
+
x
2
x
+
√
1
+
x
2
⇒
y
1
√
1
+
x
2
+
y
x
√
1
+
x
2
=
1
√
1
+
x
2
⇒
y
1
(
1
+
x
2
)
+
y
x
=
1
lf
f
(
x
)
=
x
2
x
+
a
then
f
′
(
a
)
=
Report Question
0%
4
0%
3
8
0%
3
4
0%
8
Explanation
Given,
f
(
x
)
=
x
2
x
+
a
using division rule,
f
′
(
x
)
=
(
x
+
a
)
(
2
x
)
−
x
2
(
1
)
(
x
+
a
)
2
=
x
(
2
x
+
2
a
−
x
)
(
x
+
a
)
2
=
x
(
x
+
2
a
)
(
x
+
a
)
2
∴
f
′
(
a
)
=
a
(
a
+
2
a
)
(
a
+
a
)
2
=
3
4
If
e
y
+
x
y
=
e
,
then
d
2
y
d
x
2
x
=
0
is
Report Question
0%
1
e
2
0%
e
−
1
0%
e
0%
None of these
Explanation
We have
e
y
+
x
y
=
e
Differentiating w.r.t
x
, we get
e
y
d
y
d
x
+
y
+
x
d
y
d
x
=
0
-----------(1)
Differentiating again w.r.t
x
, we get
e
y
d
2
y
d
x
2
+
e
y
(
d
y
d
x
)
2
+
2
d
y
d
x
+
x
d
2
y
d
x
2
=
0
-------(2)
Put
x
=
0
in
e
y
+
x
y
=
e
, we get
y
=
1
Putting
x
=
0
,
y
=
1
in (1) we get
e
d
y
d
x
+
1
=
0
⇒
d
y
d
x
=
−
1
e
Putting
x
=
0
,
y
=
1
,
d
y
d
x
=
−
1
e
in (2) we get
d
2
y
d
x
2
=
1
e
2
If
x
4
+
y
4
−
a
2
x
y
=
0
defines
y
implicitly as function of
x
, then
d
y
d
x
=
Report Question
0%
4
x
3
−
a
2
y
4
y
3
−
a
2
x
0%
−
(
4
x
3
−
a
2
y
4
y
3
−
a
2
x
)
0%
4
x
3
4
y
3
−
a
2
x
0%
−
4
x
3
4
y
3
−
a
2
x
Explanation
Given,
x
4
+
y
4
−
a
2
x
y
=
0
x
4
+
y
4
=
a
2
x
y
Differentiate both sides w.r.t
x
we get,
4
x
3
+
4
y
3
d
y
d
x
=
a
2
x
d
y
d
x
+
a
2
y
∴
d
y
d
x
=
a
2
y
−
4
x
3
4
y
3
−
a
2
x
=
−
(
4
x
3
−
a
2
y
4
y
3
−
a
2
x
)
lf
y
=
(
x
2
+
1
)
s
i
n
x
, then
y
′
(
0
)
is equal to
Report Question
0%
1
2
0%
e
2
0%
0
0%
3
2
Explanation
y
′
(
x
)
=
(
x
2
+
1
)
sin
x
(
cos
x
l
o
g
(
1
+
x
2
)
+
sin
x
(
2
x
1
+
x
2
)
)
y
′
(
0
)
=
1
(
0
+
0
)
=
0
If
f
(
x
)
=
e
x
g
(
x
)
,
g
(
0
)
=
1
,
g
′
(
0
)
=
3
, then
f
′
(
0
)
is
Report Question
0%
0
0%
4
0%
3
0%
7
Explanation
f
′
(
x
)
=
e
x
g
(
x
)
+
e
x
g
′
(
x
)
For x=0:
f
′
(
0
)
=
g
(
0
)
+
g
′
(
0
)
=
4
If
a
x
2
+
2
h
x
y
+
b
y
2
+
2
g
x
+
2
f
y
+
c
=
0
then
d
y
d
x
=
Report Question
0%
−
(
a
x
+
h
y
+
g
h
x
+
b
y
+
f
)
0%
−
(
a
x
+
h
y
+
g
b
x
+
h
y
+
f
)
0%
−
(
h
x
+
b
y
+
f
a
x
+
h
y
+
g
)
0%
−
(
h
x
+
b
y
+
f
h
x
+
a
y
+
g
)
Explanation
Given:
a
x
2
+
2
h
x
y
+
b
y
2
+
2
g
x
+
2
f
y
+
c
=
0
2
a
x
+
2
h
y
+
2
h
x
d
y
d
x
+
2
b
y
d
y
d
x
+
2
g
+
2
f
d
y
d
x
=
0
−
(
2
a
x
+
2
h
y
+
2
g
)
2
h
x
+
2
b
y
+
2
f
=
d
y
d
x
If
u
=
tan
−
1
(
x
2
+
y
2
x
+
y
)
,
then
x
d
u
d
x
+
y
d
u
d
y
=
Report Question
0%
sin
2
u
0%
1
2
sin
2
u
0%
1
3
sin
2
u
0%
2
sin
2
u
Explanation
Given,
u
=
tan
−
1
(
x
2
+
y
2
x
+
y
)
⇒
tan
u
=
x
2
+
y
2
(
x
+
y
)
On differentiating both the sides w.r.t.
x
and
y
respectively, we get
sec
2
u
d
u
d
x
=
2
x
(
x
+
y
)
−
(
x
2
+
y
2
)
(
x
+
y
)
2
....(1)
and
sec
2
u
d
u
d
y
=
2
y
(
x
+
y
)
−
(
x
2
+
y
2
)
(
x
+
y
)
2
....(2)
Multiplying
x
with equation (1) and
y
to equation (2).
Therefore,
x
sec
2
u
d
u
d
x
=
x
3
+
2
x
2
y
−
x
y
2
(
x
+
y
)
2
....(3)
and
y
sec
2
u
d
u
d
y
=
y
3
+
2
x
y
2
−
x
2
y
(
x
+
y
)
2
....(4)
Adding equations (3) and (4), we get
sec
2
u
(
x
d
u
d
x
+
y
d
u
d
y
)
=
x
3
+
2
x
2
y
−
x
y
2
+
y
3
+
2
x
y
2
−
x
2
y
(
x
+
y
)
2
=
x
3
+
x
2
y
+
y
3
+
x
y
2
(
x
+
y
)
2
=
(
x
+
y
)
(
x
2
+
y
2
)
(
x
+
y
)
2
=
x
2
+
y
2
x
+
y
=
tan
u
Thus
x
d
u
d
x
+
y
d
u
d
y
=
sin
u
cos
u
=
1
2
sin
2
u
If
y
=
√
sin
x
+
y
, then
d
y
d
x
is
Report Question
0%
cos
x
2
y
−
1
0%
cos
x
1
−
2
y
0%
cos
x
2
y
+
1
0%
sin
x
2
y
−
1
Explanation
We have,
y
=
√
sin
x
+
y
On squaring both the sides, we get
y
2
=
sin
(
x
)
+
y
Differentiating w.r.t.
x
,
2
y
d
y
d
x
=
cos
x
+
d
y
d
x
2
y
d
y
d
x
−
d
y
d
x
=
cos
x
d
y
d
x
(
2
y
−
1
)
=
cos
x
d
y
d
x
=
cos
x
2
y
−
1
If
√
x
+
√
y
=
4
, then find
d
x
d
y
at
y
=
1
.
Report Question
0%
3
0%
4
0%
−
3
0%
−
4
Explanation
√
x
+
√
y
=
4
Differentiating both sides of the given equation w.r.t. y, we get
1
2
√
x
d
x
d
y
+
1
2
√
y
=
0
or
d
x
d
y
=
−
√
x
√
y
=
√
y
−
4
√
y
........................................Since
√
x
+
√
y
=
4
or
[
d
x
d
y
]
y
=
1
=
1
−
4
1
=
−
3
If
x
3
+
y
3
=
3
a
x
y
, then
d
y
d
x
is
Report Question
0%
a
y
−
y
2
x
2
−
a
x
0%
a
y
−
x
2
y
2
−
a
x
0%
b
y
−
x
2
y
2
−
b
x
0%
None of these
Explanation
Given that
x
3
+
y
3
=
3
a
x
y
We have to find
d
y
d
x
Differentiate w r to
x
d
d
x
(
x
3
+
y
3
−
3
a
x
y
)
=
0
3
x
2
+
3
y
2
d
y
d
x
−
3
a
(
x
d
y
d
x
+
y
)
=
0
3
x
2
+
3
y
2
d
y
d
x
−
3
a
x
d
y
d
x
−
3
a
y
=
0
Dividing throughout by
3
d
y
d
x
(
y
2
−
a
x
)
=
a
y
−
x
2
d
y
d
x
=
a
y
−
x
2
y
2
−
a
x
For the function
f
(
x
)
=
x
100
100
+
x
99
99
+
.
.
.
.
.
.
.
.
.
.
.
+
x
2
2
+
x
+
1
,
f
′
(
1
)
=
Report Question
0%
x
100
0%
100
0%
101
0%
None of these
Explanation
G
i
v
e
n
:
d
d
x
(
x
n
)
=
n
x
n
−
1
∴
f
o
r
f
(
x
)
=
x
100
100
+
x
99
99
+
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
+
x
2
2
+
1
f
′
(
x
)
=
100
x
99
100
+
99
x
98
99
+
.
.
.
.
.
.
.
.
.
.
.
.
.
+
2
x
2
+
1
H
e
r
e
f
′
(
1
)
=
1
+
1
+
.
.
.
.
.
.
.
.
.
t
o
100
t
e
r
m
=
100
H
e
n
c
e
f
(
x
)
=
x
100
100
+
x
99
99
+
.
.
.
.
.
.
.
.
.
.
.
+
x
2
2
+
x
+
1
=
100
If
y
=
x
+
1
x
+
1
x
+
1
x
+
⋯
then
d
y
d
x
=
Report Question
0%
d
y
d
x
=
x
2
x
−
y
0%
d
y
d
x
=
y
2
y
−
x
0%
d
y
d
x
=
2
y
2
x
−
y
0%
None of these
Explanation
We have
y
=
x
+
1
x
+
1
x
+
1
x
+
⋯
=
x
+
1
y
or
y
2
=
x
y
+
1
or
2
y
d
y
d
x
=
y
+
x
d
y
d
x
+
0
[
Differentiating both sides w.r.t.
x
]
or
d
y
d
x
(
2
y
−
x
)
=
y
or
d
y
d
x
=
y
2
y
−
x
Report Question
0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct
Explanation
Let
f
(
x
)
=
(
p
x
+
q
)
(
r
x
+
s
)
We have,
d
(
u
v
)
d
x
=
u
′
v
+
u
v
′
∴
f
′
(
x
)
=
[
d
d
x
(
p
x
+
q
)
]
(
r
x
+
s
)
+
(
p
x
+
q
)
d
d
x
(
r
x
+
s
)
=
p
(
r
x
+
s
)
+
(
p
x
+
q
)
(
−
r
x
2
)
=
p
r
x
+
p
s
−
p
r
x
−
q
r
x
2
=
p
s
−
q
r
x
2
Hence, assertion is incorrect but reason is correct.
If
x
2
+
y
2
=
t
−
1
t
and
x
4
+
y
4
=
t
2
+
1
t
2
, then
x
3
y
d
y
d
x
=
Report Question
0%
0
0%
1
0%
−
1
0%
none of these
Explanation
We have
x
2
+
y
2
=
t
−
1
t
..........(1)
x
4
+
y
4
=
t
2
+
1
t
2
................(2)
On squaring equation (1), we get
(
x
2
+
y
2
)
2
=
t
2
+
1
t
2
−
2
x
4
+
2
x
2
y
2
+
y
4
=
x
4
+
y
4
−
2
..............from (2)
∴
2
x
2
y
2
=
−
2
x
2
y
2
=
−
1
y
2
=
−
1
x
2
Differentiating both the sides,
2
y
d
y
d
x
=
2
x
3
x
3
y
d
y
d
x
=
1
If
x
y
+
y
2
=
tan
x
+
y
, then
d
y
d
x
is equal to
Report Question
0%
sec
2
x
−
y
(
x
+
2
y
−
1
)
0%
cos
2
x
+
y
(
x
+
2
y
−
1
)
0%
sec
2
x
−
y
(
2
x
+
y
−
1
)
0%
cos
2
x
+
y
(
2
x
+
2
y
−
1
)
Explanation
The given relation is
x
y
+
y
2
=
tan
x
+
y
.
Differentiating both sides with respect to
x
, we get
d
d
x
(
x
y
)
+
d
d
x
(
y
2
)
=
d
d
x
(
tan
x
)
+
d
y
d
x
or
[
y
.1
+
x
.
d
y
d
x
]
+
2
y
d
y
d
x
=
sec
2
x
+
d
y
d
x
or
(
x
+
2
y
−
1
)
d
y
d
x
=
sec
2
x
−
y
∴
d
y
d
x
=
sec
2
x
−
y
(
x
+
2
y
−
1
)
If
y
x
=
x
y
, then find
d
y
d
x
.
Report Question
0%
x
(
y
log
y
−
y
)
y
(
x
log
x
−
x
)
0%
y
(
x
log
y
−
y
)
x
(
y
log
x
−
x
)
0%
y
(
x
log
y
−
y
)
0%
y
(
x
log
x
−
y
)
x
(
y
log
y
−
x
)
Explanation
y
x
=
x
y
Taking log on both sides
⇒
log
y
x
=
log
x
y
⇒
x
log
y
=
y
log
x
⇒
x
1
y
d
y
d
x
+
log
y
×
1
=
y
x
+
log
x
d
y
d
x
⇒
(
x
y
−
log
x
)
d
y
d
x
=
y
x
−
log
y
⇒
d
y
d
x
=
y
x
−
log
y
x
y
−
log
x
=
y
(
y
−
x
log
y
)
x
(
x
−
y
log
x
)
=
y
(
x
log
y
−
y
)
x
(
y
log
x
−
x
)
Differentiate
√
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
(
x
−
5
)
with respect to
x
.
Report Question
0%
1
2
√
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
(
x
−
5
)
[
1
x
−
1
−
1
x
−
2
−
1
x
−
3
−
1
x
−
4
−
1
x
−
5
]
0%
1
2
√
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
(
x
−
5
)
[
1
x
−
2
+
1
x
−
3
+
1
x
−
4
+
1
x
−
5
]
0%
1
2
√
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
(
x
−
5
)
[
1
x
−
1
+
1
x
−
2
−
1
x
−
3
−
1
x
−
4
−
1
x
−
5
]
0%
None of these
Explanation
Let
y
=
√
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
(
x
−
5
)
Taking logarithm on both sides, we get
log
y
=
l
o
g
√
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
(
x
−
5
)
=
1
2
[
log
(
x
−
1
)
+
log
(
x
−
2
)
−
log
(
x
−
3
)
−
log
(
x
−
4
)
−
log
(
x
−
5
)
]
....................(Since
log
a
b
=
log
a
−
log
b
and
log
a
b
=
log
a
+
log
b
)
Differentiating both sides w.r.t.
x
, we get
1
y
d
y
d
x
=
1
2
(
1
x
−
1
d
d
x
(
x
−
1
)
+
1
x
−
2
d
d
x
(
x
−
2
)
−
1
x
−
3
d
d
x
(
x
−
3
)
−
1
x
−
4
d
d
x
(
x
−
4
)
−
1
x
−
5
d
d
x
(
x
−
5
)
)
∴
d
y
d
x
=
1
2
√
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
−
4
)
(
x
−
5
)
[
1
x
−
1
+
1
x
−
2
−
1
x
−
3
−
1
x
−
4
−
1
x
−
5
]
If
y
=
x
−
1
2
+
log
5
x
+
sin
x
cos
x
+
2
x
, then find
d
y
d
x
Report Question
0%
−
1
2
x
−
3
/
2
+
1
x
log
e
5
+
sec
2
x
+
2
x
log
2
0%
1
2
x
−
3
/
2
+
1
x
log
e
5
+
sec
2
x
+
2
x
log
2
0%
−
3
2
x
−
3
/
2
+
1
x
log
e
5
+
sec
2
x
+
2
x
log
2
0%
−
1
2
x
−
3
/
2
+
1
x
log
e
5
+
cos
2
x
+
2
x
log
2
Explanation
Here, we have function
y
=
x
−
1
/
2
+
log
5
x
+
tan
x
+
2
x
On differentiating w.r.t x, we get
d
y
d
x
=
d
d
x
(
x
)
−
1
/
2
+
d
d
x
(
log
5
x
)
+
d
d
x
tan
x
+
d
d
x
(
2
x
)
=
−
1
2
(
x
)
−
1
/
2
−
1
+
1
x
log
e
5
+
sec
2
x
+
2
x
log
2
=
−
1
2
x
−
3
/
2
+
1
x
log
e
5
+
sec
2
x
+
2
x
log
2
If
y
=
e
x
sin
x
, then find
d
y
d
x
Report Question
0%
e
x
(
sin
x
+
cos
x
)
0%
e
x
(
sin
x
−
cos
x
)
0%
e
x
sin
x
0%
None of these
Explanation
Given that
y
=
e
x
.
sin
x
On differentiating, we get
d
y
d
x
=
sin
x
d
d
x
e
x
+
e
x
d
d
x
sin
x
d
y
d
x
=
sin
x
e
x
+
e
x
cos
x
∴
d
y
d
x
=
e
x
(
sin
x
+
cos
x
)
d
y
d
x
for
y
=
x
x
is
Report Question
0%
x
x
(
1
−
log
x
)
0%
x
x
(
1
−
log
y
)
0%
x
x
(
1
+
log
y
)
0%
x
x
(
1
+
log
x
)
Explanation
y
=
x
x
Taking
log
on both the sides
log
y
=
log
(
x
x
)
log
y
=
x
log
x
Differentiate w.r to
x
1
y
d
y
d
x
=
x
d
d
x
(
log
x
)
+
(
log
x
)
d
d
x
(
x
)
d
y
d
x
=
y
[
x
x
+
log
x
]
d
y
d
x
=
x
x
[
1
+
log
x
]
If
y
=
x
2
+
s
i
n
−
1
x
+
l
o
g
e
x
, find
d
y
d
x
Report Question
0%
d
y
d
x
=
2
x
+
1
√
1
−
x
2
+
1
x
0%
d
y
d
x
=
x
+
1
√
1
−
x
2
+
1
x
0%
d
y
d
x
=
2
x
+
1
√
1
−
x
2
−
1
x
0%
d
y
d
x
=
2
x
−
1
√
1
−
x
2
+
1
x
Explanation
y
=
x
2
+
s
i
n
−
1
+
l
o
g
e
x
On differentiating, we get
d
y
d
x
=
d
d
x
(
x
2
)
+
d
d
x
s
i
n
−
1
x
)
+
d
d
x
(
l
o
g
e
x
)
or
d
y
d
x
=
2
(
x
)
2
−
1
+
1
√
1
−
x
2
+
d
d
x
(
l
o
g
e
x
)
d
y
d
x
=
2
x
+
1
√
1
−
x
2
+
1
x
If
y
=
√
sin
x
+
√
sin
x
+
√
sin
x
+
⋯
t
o
∞
, then
d
y
d
x
is
Report Question
0%
cos
x
1
+
2
y
0%
−
sin
x
1
−
2
y
0%
cos
x
1
−
2
y
0%
cos
x
2
y
−
1
Explanation
We have
y
=
√
sin
x
+
√
sin
x
+
√
sin
x
+
⋯
t
o
∞
The given series may be written as
y
=
√
sin
x
+
y
Squaring both the sides, we get
y
2
=
sin
x
+
y
Differentiating both sides w.r.t.
x
2
y
d
y
d
x
=
cos
x
+
d
y
d
x
2
y
d
y
d
x
−
d
y
d
x
=
cos
x
d
y
d
x
(
2
y
−
1
)
=
cos
x
d
y
d
x
=
cos
x
2
y
−
1
If
y
=
e
x
tan
x
+
x
⋅
log
e
x
, then find
d
y
d
x
Report Question
0%
d
y
d
x
=
e
x
tan
x
+
(
log
x
+
1
)
0%
d
y
d
x
=
e
x
(
tan
x
+
sec
2
x
)
+
(
log
x
+
x
)
0%
d
y
d
x
=
e
x
tan
x
+
1
0%
d
y
d
x
=
e
x
(
tan
x
+
sec
2
x
)
+
(
log
x
+
1
)
Explanation
y
=
e
x
tan
x
+
x
⋅
log
e
x
On differentiating, we get
d
y
d
x
=
d
d
x
(
e
x
tan
x
)
+
d
d
x
(
x
log
x
)
=
tan
x
d
d
x
e
x
+
e
x
d
d
x
(
tan
x
)
+
x
d
d
x
(
log
x
)
+
log
x
d
d
x
x
=
e
x
⋅
tan
x
+
e
x
⋅
sec
2
x
+
1
⋅
log
x
+
x
⋅
1
x
Hence,
d
y
d
x
=
e
x
(
tan
x
+
sec
2
x
)
+
(
log
x
+
1
)
If
y
=
(
tan
x
)
(
tan
x
)
tan
x
, then find
d
y
d
x
at
x
=
π
4
.
Report Question
0%
0
0%
1
0%
−
1
0%
2
Explanation
Taking
log
on both sides, we get
log
y
=
(
tan
x
)
tan
x
log
tan
x
Again taking
log
on both the sides
log
log
y
=
[
tan
x
log
tan
x
]
+
log
log
tan
x
Differentiating w.r.t.
x
, we get
1
y
log
y
d
y
d
x
=
log
tan
x
d
d
x
(
tan
x
)
+
tan
x
d
d
x
(
log
tan
x
)
+
d
d
x
(
log
log
tan
x
)
d
y
d
x
=
log
tan
x
.
sec
2
x
+
tan
x
.
sec
2
x
tan
x
+
sec
2
x
tan
x
.
log
tan
x
d
y
d
x
=
y
log
y
.
sec
2
x
[
log
tan
x
+
1
+
1
tan
x
.
log
tan
x
]
At
x
=
π
4
,
y
=
1
and
log
y
=
0
So, putting this values in the equation of
d
y
d
x
, we get
∴
(
d
y
d
x
)
x
=
π
4
=
0
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
1
Not Answered
29
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page