# [SOLVED] How to get this 2nd differential as a function of x?

• Jun 4th 2009, 09:41 PM
tleave2000
[SOLVED] How to get this 2nd differential as a function of x?
The question is: If $\sin y + \cos y = x$ find $\frac{d^2 y}{dx^2}$ as a function of x.

I differentiated once to get $\frac{dy}{dx}=\frac{1}{\cos y - \sin y}$

Then again to get $\frac{d^2y}{dx^2}=\frac{dy}{dx}\frac{\sin y + \cos y}{(\cos y - \sin y)^2} \Rightarrow \frac{d^2y}{dx^2} =\frac{x}{(\cos y - \sin y)^3}$

The book's answer is $\frac{d^2y}{dx^2}=\pm\frac{x}{\sqrt{(2-x^2)^3}}$. So I guess that means $\cos y -\sin y =\pm\sqrt{(2-x^2)}$. But I duno why.

Did I make a mistake? Can anyone help? Thanks.
• Jun 4th 2009, 09:48 PM
mr fantastic
Quote:

Originally Posted by tleave2000
The question is: If $\sin y + \cos y = x$ find $\frac{d^2 y}{dx^2}$ as a function of x.

I differentiated once to get $\frac{dy}{dx}=\frac{1}{\cos y - \sin y}$

Then again to get $\frac{d^2y}{dx^2}=\frac{dy}{dx}\frac{\sin y + \cos y}{(\cos y - \sin y)^2} \Rightarrow \frac{d^2y}{dx^2} =\frac{x}{(\cos y - \sin y)^3}$

The book's answer is $\frac{d^2y}{dx^2}=\pm\frac{x}{\sqrt{(2-x^2)^3}}$. So I guess that means $\cos y -\sin y =\pm\sqrt{(2-x^2)}$. But I duno why.

Did I make a mistake? Can anyone help? Thanks.

Note that $x = \cos y + \sin y \Rightarrow x^2 = 1 + 2 \cos y \sin y \Rightarrow x^2 - 1 = 2 \cos y \sin y$.

Therefore $(\cos y - \sin y )^2 = 1 - 2 \cos y \sin y = 1 - (x^2 - 1) = 2 - x^2$.
• Jun 4th 2009, 10:10 PM
tleave2000
Wow, that's pretty difficult to spot...
Thanks alot.