y = xxcosx and z =

x2+1

x2−1

Consider y = xxcosx
Taking log on both sides,
log y = log (xxcosx)
log y = x cos x log x
Differentiating with respect to x,

1

y

dy

dx

= (x cos x)

1

x

+ log x

d

dx

+ log x

d

dx

(x cos x)

1

y

dy

dx

= cos x + log x (cos x - x sin x)

dy

dx

= y (cos x + log x [cos x - x sin x)]

dy

dx

= xxcosx [cos x + log x (cos x - x sin x)] … (1)
Consider z =

x2+1

x2−1

Differentiating with respect to x,

dz

dx

=
=

(x2−1)(2x)−(x2+1)(2x)

(x2−1)2

=

2x3−2x−2x3−2x

(x2−1)2

=

−4x

(x2−1)2

... (2)
Adding (1) and (2):

d

dx

{xxcosx+

x2+1

x2−1

} =

dy

dx

+

dz

dx

= xxcosx [cos x + log x (cos x – x sin x)] –

x

(x2−1)2

OR
x = a(θ - sinθ) , y = a(1 + cosθ)
Differentiating x and y w.r.t. θ,

dx

dθ

= a (1 - cos θ) ... (1)

dy

dθ

= - a sin θ ... (2)
Dividing (2) by (1),

( dy dθ )

( dx dθ )

=

−asinθ

a(1−cosθ)

⇒

dy

dx

=

−sinθ

1−cosθ

⇒

dy

dx

=

−2sin θ 2 cos θ 2

2sin2 θ 2

⇒

dy

dx

=

−cos θ 2

sin θ 2

⇒

dy

dx

= = - cot

θ

2

... (3)
Differentiating w.r.t. x,

d

dx

(

dy

dx

) =

d

dθ

(

dy

dx

)×

dθ

dx

⇒

d2y

dx2

=

d

dθ

(

dy

dx

)×

dθ

dx

⇒

d2y

dx2

=

d

dθ

(−cot

θ

2

)×

dθ

dx

[from equation (3)]

d2y

dx2

= - (−cosec2

θ

2

×

1

2

)×

dθ

dx

=

1

2

cosec2

θ

2

×

1

( dx dθ )

=

1

2

cosec2

θ

2

×

1

a(1−cosθ)

... [from equation (1)]
=

cosec2 θ 2

2a(1−cosθ)

=

cosec2 θ 2

2a(2sin2 θ 2 )

=

1

4a

×cosec2

θ

2