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

You have
$$\frac{dy}{dx}=\frac{1}{\cos y-\sin y}.$$
Take the derivative of both sides using one of the  derivative rules:
$$\frac{d^2y}{dx^2}=\frac{(1)'(\cos y - \sin y)-(1)(\cos y-\sin y)'}{(\cos y-\sin y)^2}=?,$$
$$\mathrm{or},\quad (x^{-1})'=-x^{-2}\implies\frac{d^2y}{dx^2}=\frac{-1}{(\cos y-\sin y)^2}\cdot(\cos y-\sin y)'=?$$
Above are the beginnings to (i) the quotient rule and (ii) the power rule and chain rules.
