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
 A: 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.
A: Just differentiate both sides of ${dy\over dx}=(\cos y-\sin y)^{-1}$ with respect to $x$. This leads to:
$${d^2y\over dx^2}={d\over dx}{dy\over dx}={d\over dx}(\cos y-\sin y)^{-1}=-(\cos y-\sin y)^{-2}\cdot\Bigl[ (-\sin y){dy\over dx} -(\cos y){dy\over dx}\Bigr].$$
A: x=sin y+cos y
first derivative by implicit diffferentiation;
1=cos y(y')-sin y(y')
1=y'(cos y-sin y)
y'=1/(cos y-sin y)
second derivetive
y''=-(-sin y(y')-cos y(y'))/(cos y-sin y)^2
y"=(y'sin y+y'cos y)/(cos y-sin y)^2
subsitute the value of y'
we hwve;
y"=(1/(cos y-sin y)(sin y+cos y))/(cos y-sin y)^2
y"=(sin y+cos y)/(cos y-sin y)^2
simplifying this we have;
y"=(sin y+cosy )/(-sin 2y) or
y"=-(sin y+cos y)/2sinycosy
y"=-sin y/2sinycosy  -  cos y/2sinycosy
y"=-secy/2    -      cscy/2            ans:
