The given function is (cosx)y = (cosy)x
Taking logarithm on both the sides, we obtain
ylog cosx = xlog cosy
Differentiating both sides, we obtain
log cosx ×

dy

dx

+ y ×

d

dx

(log cos x) = log cos y ×

d

dx

(x) + x ×

d

dx

(log cos y)
⇒ log cos x ×

dy

dx

+ y ×

1

cosx

×

d

dx

(cos x) = log cos y × 1 + x ×

1

cosy

×

d

dx

(cos y)
⇒ log cos x ×

dy

dx

+

y

cosx

(- sin x) = log cos y +

x

cosy

× (- sin y) ×

dy

dx

⇒ log cos x ×

dy

dx

- y tan x - log cos y - x tan y ×

dy

dx

⇒ log cos x ×

dy

dx

+ x tan y ×

dy

dx

= log cos y + y tan x
⇒ (log cos x + x tan y) ×

dy

dx

= log cos y + y tan x
∴

dy

dx

=

logcosy+ytanx

logcosx+xtany

OR
We have,
siny = x sin (a + y)
⇒ x =

siny

sin(a+y)

Differentiating the above function we have,
1 =
⇒ sin2 (a + y) = [sin (a + y) × cos y - sin y cos (a + y)]

dy

dx

⇒ =

dy

dx

⇒

sin2(a+y)

sin(a+y−y)

=

dy

dx

⇒

sin2(a+y)

sina

=

dy

dx

⇒

dy

dx

=

sin2(a+y)

sina