in dy/dx=d(x^2+y^2) /dx, why d(t^2)/dx =2t.dt/dx
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Answered by
0
Answer:
For parametric form of equation,
d
y
d
x
=
d
y
d
t
d
x
d
t
.
Here as
x
=
t
2
−
2
t
,
d
x
d
t
=
2
t
−
2
=
2
(
t
−
1
)
and as
y
=
t
4
−
4
t
,
d
y
d
t
=
4
t
3
−
4
=
4
(
t
3
−
1
)
=
4
(
t
−
1
)
(
t
2
+
t
+
1
)
Hence
d
y
d
x
=
4
(
t
−
1
)
(
t
2
+
t
+
1
)
2
(
t
−
1
)
=
2
(
t
2
+
t
+
1
)
Answered by
1
Answer:
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