clarisse Posted January 18, 2006 Share Posted January 18, 2006 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! Link to comment Share on other sites More sharing options...
the tree Posted January 18, 2006 Share Posted January 18, 2006 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] Link to comment Share on other sites More sharing options...
JustStuit Posted January 18, 2006 Share Posted January 18, 2006 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. Link to comment Share on other sites More sharing options...
the tree Posted January 18, 2006 Share Posted January 18, 2006 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. Link to comment Share on other sites More sharing options...
JustStuit Posted January 18, 2006 Share Posted January 18, 2006 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. Link to comment Share on other sites More sharing options...
JustStuit Posted January 18, 2006 Share Posted January 18, 2006 It is confusing when people don't use the latex. I'm trying to learn how to use it more. Link to comment Share on other sites More sharing options...
clarisse Posted January 18, 2006 Author Share Posted January 18, 2006 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? Link to comment Share on other sites More sharing options...
JustStuit Posted January 18, 2006 Share Posted January 18, 2006 I didn't learn latex until like yesterday so no problem. I still don't know it well. Link to comment Share on other sites More sharing options...
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