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Mathematics
Problem
For the parametric equations
x
=
t
2
+
1
x=t^{2}+1
x
=
t
2
+
1
and
y
=
t
3
−
3
t
y=t^{3}-3t
y
=
t
3
−
3
t
, the value of
d
y
d
x
\dfrac{dy}{dx}
d
x
d
y
at
t
=
1
t=1
t
=
1
is a)
1
1
1
b)
3
2
\dfrac{3}{2}
2
3
c)
0
0
0
d)
3
3
3
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