# differentiation 4 problem

• November 22nd 2009, 10:04 PM
furor celtica
differentiation 4 problem
find d^2/dx^2 as a function of x if siny + cosy = x
i tried to solve this problem this way
differentiating once makes cos y (dy/dx) - sin y (dy/dx) = 1
dy/dx (cos y - siny) = 1
differentiating this makes d^2y/dx^2 (cosy - siny) -dy/dx (sin y + cos y) = 0
d^2y/dx^2 (cos y -siny) - x(dy/dx) = 0
from there i isolate d^2y/dx^2, but apparently its wrong
can anyone help?
• November 23rd 2009, 12:42 AM
Hello furor celtica
Quote:

Originally Posted by furor celtica
find d^2/dx^2 as a function of x if siny + cosy = x
i tried to solve this problem this way
differentiating once makes cos y (dy/dx) - sin y (dy/dx) = 1
dy/dx (cos y - siny) = 1
differentiating this makes d^2y/dx^2 (cosy - siny) -dy/dx (sin y + cos y) = 0
d^2y/dx^2 (cos y -siny) - x(dy/dx) = 0
from there i isolate d^2y/dx^2, but apparently its wrong
can anyone help?

$\frac{dy}{dx}(\cos y - \sin y )=1$

$\Rightarrow \frac{d^2y}{dx^2}(\cos y - \sin y) - \frac{dy}{dx}(\sin y + \cos y)\frac{dy}{dx}=0$

Can you continue?

(By the way, it would be much easier to read your work in LaTeX.)

PS Had you thought of expressing $\sin y + \cos y$ as $\sqrt2\sin(y+\pi/4)$? That way, you can get an explicit equation for $y$ in terms of $\arcsin x$, which will probably make the final result easier to express in terms of $x$.
• November 23rd 2009, 02:43 AM
furor celtica
can you explain that part i got wrong? i dont get it. and isnt it superfluous to express siny + cosy as a compound angle sine when it is given 'siny + cosy = x'?
• November 23rd 2009, 03:01 AM
Quote:

Originally Posted by furor celtica
can you explain that part i got wrong? i dont get it. and isnt it superfluous to express siny + cosy as a compound angle sine when it is given 'siny + cosy = x'?

When you differentiated the second term of the product $\frac{dy}{dx}(\cos y - \sin y)$, you wrote $-(\sin y + \cos y)$, whereas it should be (as I wrote) $-(\sin y +\cos y)\frac{dy}{dx}$, because you're differentiating with respect to $x$, not $y$.

$\sin y + \cos y = x$

$\Rightarrow \sqrt2\sin(y+\pi/4)=x$

$\Rightarrow \sin(y+\pi/4)=\frac{x}{\sqrt2}$

$\Rightarrow y+\pi/4 = \arcsin\left(\frac{x}{\sqrt2}\right)$

$\Rightarrow y= \arcsin\left(\frac{x}{\sqrt2}\right)-\pi/4$

Differentiate this twice, and you're done.