I've been trying to work out this problem but I can't do it!

It says:

 

Find (d squared y)/(dx squared) as a function of x if sin y + cos y =x

 

I try: cosy(dy/dx) - siny(dy/dx)=1

(dy/dx)(cosy-siny)=1

dy/dx=1/(cosy-siny)

 

well after that I've tried many things but none of them appear to work...

so... could anybody help me please!! thank you very much!

I'm not sure but I think there are two basic generalisations you should be using:

 

[math]\frac{d}{dx}\sin x=\cos x[/math] and [math]\frac{d}{dx}\cos x=-\sin x[/math]

 

ergo

 

if [math]f(y)=\sin{y}+\cos{y}[/math]

 

then [math]f'(y)=\cos{y}-\sin{y}[/math]

 

and [math]f''(y)=-\sin{y}-\cos{y}[/math]

For derivatives they usually use dy/dx or d/dx or with whatever variable, what is up with the squared parts of it?

 

Find (d squared y)/(dx squared)

 

Maybe we haven't got there in my calc class.

For derivatives they usually use dy/dx or d/dx or with whatever variable, what is up with the squared parts of it?

[math]\frac{d^{2}y}{dx^2}[/math] is the second deriative, the deriative of the deriative.

Oh I misunderstood. Most of the time we use [math]

f'(x) [/math] or [math] f''(x) [/math] but we use [math]

\frac{d^{2}y}{dx^2}

[/math] also.

It is confusing when people don't use the latex. I'm trying to learn how to use it more.

Hm sorry for the confusion JustStuit, I haven't taken the time to learn to use latex but oh well..

 

I think the answer's like this but I could be wrong:

I'll write the second derivative as d2y/dx2...

 

siny+cosy=x

cosy(dy/dx)-siny(dy/dx)=1

dy/dx = 17(cosy-siny) = (cosy-siny)^-1

d2y/dx2= (-1(cosy-siny)^-2) * (-siny(dy/dx) - cosy(dy/dx)

d2y/dx2= (siny+cosy)/(cosy-siny)^3

(cosy+siny)^2 = (siny)^2 + (cosy)^2 + 2cosysiny = 1 + 2cosysiny

 

x^2 - 1 = 2cosysiny

(cosy-siny)^2=1-2cosysiny

cosy-siny= (1-((x^2)-1)^1/2) = (2-x^2)^1/2

 

so now

 

x/((2-x^2)^1/2)^3 = x/((2-x^2)^3/2)

 

Is this right?

I didn't learn latex until like yesterday so no problem. I still don't know it well.