UPSC NDA Math Model Paper 3

Given: y=sin−1(sin2x)
To find: Slope of tangent at x = 0
y=sin−1(sin2x)
Differentiating to both sides w.r.t x we get,
⇒

dy

dx

=

1

√1−sin4x

×

d(sin2x)

dx

⇒

dy

dx

=

1

√1−sin2x

×2‌sin‌x×cos‌x
So, calculating

dy

dx

at x=0.

dy

dxx=0

=

1

√1−0

×2×0×1
⇒

dy

dxx=0

−0
Hence, the slope of the tangent to the given curve is 0.