How to find the second derivative (with respect to $x$) of $\cos y + \sin y = x$?

Question

How to find the second derivative (with respect to $x$) of $\cos y + \sin y = x$? The answer is $$\pm \dfrac {x}{(2-x^2)^{3/2}}$$

I dont understand how it goes from using just $\cos x$ and $\sin x$ to simple terms and a root. I've tried multiple times and have not gotten anywhere near the answer.

Thanks.

Answer 1

If you square both sides of the equation you get

$$\sin y+\cos y=x\implies\sin (2y)=x^2-1\implies y=\frac12\arcsin(x^2-1)$$

Now differentiate the last expression and find if you are missing some solution of the original equation. Note that $x\in[-\sqrt 2,\sqrt 2]$.

Answer 2

In this case you can write $y$ explicitly in terms of $x$... up to a choice of branch for the $\arcsin$ function.

$$ \sin y + \cos y = x \Rightarrow (\sin y + \cos y)^2 = x ^2 \Leftrightarrow \\ \sin^2 y + 2 \sin y \cos y + \cos^2 y = x^2 \Leftrightarrow\\ \sin(2y) = x^2-1 \Leftrightarrow \\ y = \frac 12 \arcsin(x^2-1) $$

Hence,

$$ y'(x) = \pm \frac 12 \frac{2x}{\sqrt{1-(x^2-1)^2}}=\pm x (1-(x^2-1)^2)^{-1/2} $$

Can you proceed?

Note that the relation does not globally define $y$ as a function of $x$... This is why we have the $\pm$, the "graph" would be a periodic function about the $y$ axis and when you pick a branch the function can be increasing or decreasing.