Question

Find y'' for siny=cos2x?

differentiated first stage giving me (2sin2x)/cosy

Answer 1

`(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) }`