Find y'' for siny=cos2x?

differentiated first stage giving me (2sin2x)/cosy

differentiated first stage giving me (2sin2x)/cosy

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Steve M Share
Mar 21, 2018

Answer:

# (d^2y)/(dx^2) = 4{ ( sin^2 2x \ siny - cos^2y \ cos2x ) / (cos^3y) } #

Explanation:

Given that:

# siny = cos2x#

We can differentiate Implicitly to get:

# cosydy/dx = -2sin2x #

# :. dy/dx = -2(sin2x)/cosy #

And differentiating again, we get:

# (d^2y)/(dx^2) = -2{ ( (cosy)(2cos2x) - (sin2x)(-sinydy/dx) ) / (cosy)^2} #

# \ \ \ \ \ \ \ = -2{ ( (cos^2y)(2cos2x)/cosy + (sin2x)(siny(-2(sin2x)/cosy)) ) / (cosy)^2} #

# \ \ \ \ \ \ \ = -4{ ( cos^2y \ cos2x - sin^2 2x \ siny ) / (cos^3y) } #

# \ \ \ \ \ \ \ = 4{ ( sin^2 2x \ siny - cos^2y \ cos2x ) / (cos^3y) } #

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