Let #siny + cosx = 1# then find #(d^2y)/(dx^2)#?

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Answer 1 · 2018-05-30T12:54:37

# (d^2y)/(dx^2) = siny/cosy \ sin^2x/cos^2y + cosx /cosy #

We have:

# siny+cosx=1 #

If we differentiate (implicitly) wrt #x# we get:

# cosy \ dy/dx -sinx = 0 # .... [A]

If we differentiate (implicitly) wrt #x# again whilst applying the product rule we get:

# (cosy) \ (d^2y)/(dx^2) -(siny \ dy/dx) \ dy/dx - cosx = 0 #

# :. (d^2y)/(dx^2) -siny/cosy \ (dy/dx)^2 - cosx/cosy = 0 #

And, from [A] we get #dy/dx = sinx/cosy#, so substituting this result in the above we find that:

# (d^2y)/(dx^2) -siny/cosy \ (sinx/cosy)^2 - cosx /cosy= 0 #

# :. (d^2y)/(dx^2) = siny/cosy \ sin^2x/cos^2y + cosx /cosy #