$ x\cos y + y \cos x = \pi$ , how do I find $y''(0)$ in easiest way possible? I considered brute force differentiating but that is very hard. I also tried expanding the cosines and '$y$' as a Taylor series but I  don't think that helps much either
 A: Differentiating both sides with respect to $x$, we obtain \begin{equation*}\cos y - x\sin y\frac{\mathop{}\!\mathrm{d}y}{\mathop{}\!\mathrm{d}x}+\frac{\mathop{}\!\mathrm{d}y}{\mathop{}\!\mathrm{d}x}\cos x-y\sin x=0\end{equation*} i.e. \begin{equation*}(\cos y-y\sin x)+(\cos x-x\sin y)\frac{\mathop{}\!\mathrm{d}y}{\mathop{}\!\mathrm{d}x}=0 \qquad (*)\end{equation*} When $x=0$, we have $0+y(1)=\pi$, so $y=\pi$. Thus $(-1-0)+\left(1-0\right)\frac{\mathop{}\!\mathrm{d}y}{\mathop{}\!\mathrm{d}x}=0$, so $\frac{\mathop{}\!\mathrm{d}y}{\mathop{}\!\mathrm{d}x}=1$ when $x=0$.
Now, differentiating $(*)$ again gives \begin{equation*}-\sin y\frac{\mathop{}\!\mathrm{d}y}{\mathop{}\!\mathrm{d}x} - \frac{\mathop{}\!\mathrm{d}y}{\mathop{}\!\mathrm{d}x}\sin x - y\cos x +\left(-\sin x-\sin y-x\cos y\frac{\mathop{}\!\mathrm{d}y}{\mathop{}\!\mathrm{d}x}\right)\frac{\mathop{}\!\mathrm{d}y}{\mathop{}\!\mathrm{d}x} + \left(\cos x - x\sin y\right)\frac{\mathop{}\!\mathrm{d}^{2}y}{\mathop{}\!\mathrm{d}x^{2}} = 0\end{equation*} Substituting the values from above, we obtain $0-0-\pi(1)+(0-0-0)(1)+\left(1-0\right)\frac{\mathop{}\!\mathrm{d}^{2}y}{\mathop{}\!\mathrm{d}x^{2}} = 0$, so \begin{equation*}\left.\frac{\mathop{}\!\mathrm{d}^{2}y}{\mathop{}\!\mathrm{d}x^{2}}\vphantom{\int}\right\rvert_{x=0}^{} = \pi\end{equation*}
A: Use the implicit function theorem for
$$f=x \cos(y)+y \cos(x)-\pi=0$$
$$f'_x=\cos(y)-y \sin(x)\qquad \qquad f'_y=-x \sin(y)+\cos(x)$$
$$y'=\frac{dy}{dx}=-\frac{f'_x}{f'_y}=-\frac{\cos(y)-y \sin(x) }{-x \sin(y)+\cos(x) }$$ and now differentiate with respect to $x$; in the rhs, you will have some $y'$.
A: Since $F(x,y(x)) = x\cos y + y\cos x$ is differentiable, we can use the implicit differentiation formula:
\begin{align*} y' =\frac{dy}{dx} & = - \frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}\\
& = -\frac{\cos y -y\sin x}{\cos x - x\sin y}.
\end{align*}
Now we have
$$y'\cos x - y'x\sin y -y\sin x + \cos y  = 0.$$
Differentiate with respect to $x$ on both sides and simplify .
A: I concur with @Prasiortle. Since $\cos y-xy^\prime\sin y-y\sin x+y^\prime\cos x=0$,$$-2y^\prime\sin y-x(y^{\prime\prime}\sin y+y^{\prime2}\cos y)-y\cos x-2y^\prime\sin x+y^{\prime\prime}\cos x=0,$$i.e.$$\begin{align}y^\prime&=\frac{\cos y-y\sin x}{x\sin y-\cos x},\\y^{\prime\prime}&=\frac{xy^{\prime2}\cos y+2y^\prime(\sin x+\sin y)+y\cos x}{\cos x-x\sin y}.\end{align}$$At $x=0$ the original equation gives $y=\pi$, while the above results give $y^\prime=-\cos y=1$ and $y^{\prime\prime}=2y^\prime\sin y+y=\pi$.
A: $x\cos y + y\cos x = \pi$
What is $y(0)$?
$0\cos (y(0)) + y(0)\cos 0 = \pi\
y(0) = \pi\
What is $y'$?
$\cos y -y\sin x + (-x\sin y + \cos x)y' = 0\\
y'= \frac {y\sin x - \cos y}{\cos x - x\sin y} = \frac {u}{v}\\
$
And $y'(0)$?
$u(0) = \pi \sin 0 - \cos \pi = 1\\
v(0) = \cos 0 - 0\sin \pi = 1\\
y'(0) = 1$
$y'' = \frac {u'v - uv'}{v^2}\\
u' = y\cos x + (\sin x + \sin y) y'\\
u'(0) = \pi\\
v' = -sin x - sin y + (-x\cos y) y'\\
v'(0) = 0\\
y''(0) = \pi$
A: Well well, it's been over six months and I had recently found a better way to do this.
$$ x \cos y + y \cos x = \pi$$
Now, rewrite as:
$$ Q(x,y) = x \cos y + y \cos x - \pi$$
r
D.w.r.t.x and use implicit function theorem
$$ \frac{dQ}{dx} = \left[  \cos y  - y \sin x \right] +y' \left [ -x \sin y + \cos x \right] $$
Now notice that $\frac{dQ}{dx} = G(x,y,y')$ , use the implicit function theorem yet again:
$$ \frac{dG}{dx} =( \left[ - y \cos x \right] + y' \left[-\sin y - \sin x \right] )+ (\left[ - \sin y - \sin x \right] + y' \left[ -x \cos y \right])y' +\left[ (-x \sin y + \cos x) \right] y'' $$
Hence, our three implicit equations relating the function and it's derivatives are:
$$ \left[  x \cos y + y \cos x - \pi \right]=0 \tag{0}$$
$$ \left[  \cos y  - y \sin x \right] +y' \left [ -x \sin y + \cos x \right] =0\tag{1} $$
$$( \left[ - y \cos x \right] + y' \left[-\sin y - \sin x \right] )+ (\left[ - \sin y - \sin x \right] + y' \left[ -x \cos y \right])y' +\left[ (-x \sin y + \cos x) \right] y'' =0 \tag{2}$$
Plugging $ x = 0$,
$$ y- \pi = 0 \tag{0}$$
$$ \left[ \cos(y) \right] + y' =0$$
$$ ( -y -y' \sin y) + (- \sin y  )y' + \left[ 1 \right] y''=0$$
Simplfying,
$$ y= \pi$$
$$ y'=1$$
$$y''=\pi$$
Refer here
