# 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

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+1 for showing you've worked on the problem. –  David Mitra Nov 29 '11 at 20:28

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.

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

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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:

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