Find the differential dy for y = (cos^2)x

y = ( cos x ) ²

dy/dx = 2 cos x (- sin x )

dy/dx = - 2 sin x cos x

dy/dx = - sin 2x

Thank you ---pleased to help.

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your question is the same as

y = (cos x)^2

so just use ln to solve this....then differentiate as usual..
ln y = 2 . ln cos x

1/y . dy/dx = uv' +u'v

u = 2 v = ln cos x
u' = 0 v' = (1/cos x) . -sin x . 1

therefore :
1/y . dy/dx = (-2sin x)/cos x

dy/dx = y[ (-2sin x)/cos x ]

y = cos²x
dy / dx = -2sinxcosx
dy = -2sinxcosx dx
dy = -sin(2x) dx