Let $sin y + cos x = 1$ $siny + cosx = 1$ then find $\frac{d^{2} y}{{d x}^{2}}$ $(d^2y)/(dx^2)$ ?

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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:

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

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

Let siny + cosx = 1siny+cosx=1siny + cosx = 1 then find (d^2y)/(dx^2)d2ydx2(d^2y)/(dx^2)?