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Byju's Answer
Standard XII
Mathematics
Parametric Differentiation
`A function i...
Question
`A function is represented parametrically by the equation
x
=
1
+
t
t
3
,
y
=
3
2
t
2
+
2
t
then
d
y
d
x
−
x
(
d
y
d
x
)
2
has the value equal to
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Solution
x
=
1
+
t
t
3
,
y
=
3
2
t
2
+
2
t
d
x
d
t
=
d
d
t
[
1
t
2
+
1
t
2
]
d
x
d
t
=
[
−
3
t
4
−
2
t
3
]
d
2
x
d
t
2
=
[
12
t
5
+
6
t
4
]
d
y
d
t
=
−
3
t
3
−
2
t
2
=
−
[
3
+
2
t
t
3
]
d
2
y
d
t
−
9
t
4
+
4
t
3
=
[
9
+
4
t
t
4
]
d
y
d
x
−
x
(
d
y
d
x
)
2
=
?
d
y
d
x
=
d
y
d
t
d
x
d
t
=
−
(
3
+
2
t
t
3
)
−
(
3
+
2
t
t
4
)
=
t
d
y
d
x
−
x
(
d
y
d
x
)
2
=
t
−
(
t
+
t
)
t
3
×
t
2
=
t
−
(
1
+
t
)
t
d
y
d
x
−
x
(
d
y
d
x
)
2
=
t
−
1
t
−
1
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Similar questions
Q.
If
x
=
1
+
t
t
3
,
y
=
3
2
t
2
+
2
t
, then
x
(
d
y
d
x
)
3
−
d
y
d
x
is equal to
Q.
If a function is represented parametrically by
x
=
2
t
2
+
5
t
and
y
=
8
t
2
+
15
t
3
6
t
5
, then the value of
∣
∣
∣
x
d
y
d
x
−
d
y
d
x
∣
∣
∣
at
t
=
−
1
is
Q.
A function is represented parametrically by the equations
x
=
log
|
2
t
|
,
y
=
∣
∣
tan
−
1
|
t
|
∣
∣
,
t
<
0.
Then the value of
∣
∣ ∣
∣
(
1
+
t
2
)
2
[
1
t
d
2
y
d
x
2
−
(
d
y
d
x
)
2
]
∣
∣ ∣
∣
is
Q.
If a function is represented parametrically by
x
=
2
t
2
+
5
t
and
y
=
8
t
2
+
15
t
3
6
t
5
, then the value of
∣
∣
∣
x
d
y
d
x
−
d
y
d
x
∣
∣
∣
at
t
=
−
1
is
Q.
If a curve is represented parametrically by the equations
x
=
f
(
t
)
and
y
=
g
(
t
)
,
then
(
d
2
y
d
x
2
)
(
d
2
x
d
y
2
)
is equal to
(
where
f
′
(
t
)
≠
0
and
g
′
(
t
)
≠
0
)
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