Second derivative using implicit differentiation with respect to $x$ of $x = \sin y + \cos y$ I am running into trouble with this question:
I get as far as
$$1 = \cos y\frac{dy}{dx} - \sin y\frac{dy}{dx}$$
$$1 = \frac{dy}{dx} (\cos y - \sin y)$$
$$\frac{dy}{dx} = \frac{1}{\cos y-\sin y}$$
Second derivative:
Unsure how to continue here

Just differentiate both sides of ${dy\over dx}=(\cos y-\sin y)^{-1}$ with respect to $x$. This leads to:
$${d^2y\over dx^2}={d\over dx}{dy\over dx}={d\over dx}(\cos y-\sin y)^{-1}=-(\cos y-\sin y)^{-2}\cdot\Bigl[ (-\sin y){dy\over dx} -(\cos y){dy\over dx}\Bigr].$$
