In order to find derivative of function in such form, we have by chain rule dy
dt  = dy dx
dx dt
⋅
or
dy
dx  =
whenever
0
dy
dx
dt
dx
dt
dt


≠




Thus
dy
dx  =
( )
as
( ) and
( )
g t( )
dy
dx
g t
f t
f t
dt
dt
′


=
′
=
′


′

  [provided f′(t) ≠ 0]
Example 31 Find dy
dx , if x = a cos θ, y = a sin θ Solution Given that
x = a cos θ, y = a sin θ
Therefore
dx
dθ  = – a sin θ, dy
dθ  = a cos θ
Hence
dy
dx  =
cos
cot
sin
dy
a
d
dx
a
d
θ
θ =
= −
θ
−
θ
θ
Example 32 Find dy
dx
, if x = at2, y = 2at Solution Given that x = at2, y = 2at
So
dx
dt  = 2at   and   dy
dt = 2a
Therefore
dy
dx  =
2
1
2
dy
a
dt
dx
at
t
dt
=
=
Rationalised 2023-24
 MATHEMATICS
136
Example 33 Find dy
dx
, if x = a (θ + sin θ), y = a (1 – cos θ)