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Byju's Answer
Standard XII
Mathematics
Evaluation of a Determinant
If y=logcos x...
Question
If
y
=
log
cos
x
sin
x
log
sin
x
cos
x
-
1
+
sin
-
1
2
x
1
+
x
2
,
find
d
y
d
x
a
t
x
=
π
4
Open in App
Solution
We
have
,
y
=
log
cos
x
sin
x
log
sin
x
cos
x
-
1
+
sin
-
1
2
x
1
+
x
2
⇒
y
=
log
cos
x
sin
x
log
cos
x
sin
x
+
sin
-
1
2
x
1
+
x
2
∵
log
a
b
=
log
b
a
-
1
⇒
y
=
log
sin
x
log
cos
x
2
+
sin
-
1
2
x
1
+
x
2
∵
log
a
b
=
log
b
log
a
Differentiating with respect to x,
d
y
d
x
=
d
d
x
log
sin
x
log
cos
x
2
+
d
d
x
sin
-
1
2
x
1
+
x
2
⇒
d
y
d
x
=
2
log
sin
x
log
cos
x
d
d
x
log
sin
x
log
cos
x
+
1
1
-
2
x
1
+
x
2
2
×
d
d
x
2
x
1
+
x
2
⇒
d
y
d
x
=
2
log
sin
x
log
cos
x
log
cos
x
d
d
x
log
sin
x
-
log
sin
x
d
d
x
log
cos
x
log
cos
x
2
+
1
+
x
2
1
+
x
4
-
2
x
2
1
+
x
2
2
-
2
x
2
x
1
+
x
2
2
⇒
d
y
d
x
=
2
log
sin
x
log
cos
x
log
cos
x
×
1
sin
x
d
d
x
sin
x
-
log
sin
x
×
1
cos
x
d
d
x
cos
x
log
cos
x
2
+
1
+
x
2
1
+
x
4
-
2
x
2
1
+
x
2
2
-
2
x
2
x
1
+
x
2
2
⇒
d
y
d
x
=
2
log
sin
x
log
cos
x
log
cos
x
×
cos
x
sin
x
+
log
sin
x
×
sin
x
cos
x
log
cos
x
2
+
1
+
x
2
1
-
x
2
2
2
+
2
x
2
-
4
x
2
1
+
x
2
2
⇒
d
y
d
x
=
2
log
sin
x
log
cos
x
3
cot
x
log
cos
x
+
tan
x
log
sin
x
+
2
1
+
x
2
put
x
=
π
4
⇒
d
y
d
x
=
2
log
sin
π
4
log
cos
π
4
3
cot
π
4
log
cos
π
4
+
tan
π
4
log
sin
π
4
+
2
1
1
+
π
4
2
⇒
d
y
d
x
=
2
1
log
1
2
2
1
×
log
1
2
+
1
×
log
1
2
+
2
16
16
+
π
2
⇒
d
y
d
x
=
2
×
2
log
1
2
log
1
2
2
+
32
16
+
π
2
⇒
d
y
d
x
=
4
1
log
1
2
+
32
16
+
π
2
⇒
d
y
d
x
=
4
1
-
1
2
log
2
+
32
16
+
π
2
⇒
d
y
d
x
=
-
8
log
2
+
32
16
+
π
2
So
,
d
y
d
x
x
=
π
4
=
8
4
16
+
π
2
-
1
log
2
Suggest Corrections
0
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