How Do I Solve a Second Derivative Using Implicit Differentiation?

[GregA]

Need help with implicit differentiation

I have only just been introduced to implicit differentiation and am cluelessly stuck on this question:
Express d^2y/dx^2 as a function of x if siny + cosy = x
my first attempt was just to simply differentiate each term, and ended up with -(siny+cosy)d^2y/dx^2 = 0. The answer in the back of the book is: x/(2-x^2)^(3/2)...which unlike my answer is a function of x...my problem is that I really don't know how to arrive at this answer.
I can't think of any trig identity that can help me here and so my best guess is that I should find the three sides of a triangle with which siny and cosy make x, but I have no idea about how to go about finding these sides with the information I have been given.
part of the denominator: sqrt(2-x^2) seems like the hypoteneuse but I am having difficulty coming up with the other two sides and to make x. I apologise if solving this should be a no-brainer but I simply don't know how to proceed. Can someone please point me in the right direction?

 

[HallsofIvy]

Then I suspect that the answer in the back of the book is wrong!

sin y+ cos y= x so, by implicit differentiation, cos y y'- sin y y'= 1 or
y'(cos y- sin y)= 1.

Differentiate again: y" (cos y- sin y)+ y'(-sin y- cosy)= 0 which is the same as y" (cos y- sin y)= (sin y+ cos y)y'= xy' (since sin y+ cos y= x).

But y'(cos - sin y)= 1 means that y'= 1/(cosy -sin y) so

y"(cos y- sin y)= x/(cos y- sin y)

y"= x/(cos y- sin y)²= x/(cos<sup>2[/sup y- 2sin ycosy+ cos2 y)= x/(1- 2sin y cos y)

Now, we know that x= sin y+ cos y so x2= (sin y+ cos y)2= sin2+ 2sin y cos y+ cos2 y= 1+ 2 sin y cos y.
1- 2 siny cos y= 2- (1+ 2 sin y cos y)= 2- x2.

The correct answer is just $$y"= \frac{x}{2-x^2}$$. There is no "3/2" power.</sup>

 

[GregA]

That was brilliant HallsofIvy...Thankyou

 

[christodoulou Andrea]

The curve y1 is y1=ax^2+ax+b.The curve y2 is y2=cx-x^2.They both have
common tangent the line which passes through the ponit (1,0).Find the a,b,c.
Please somebody help me.Thank you.

 

[HallsofIvy]

1. Do not "hijack" someone else's thread to ask a completely new question- start your own thread.

2. Show us what you have done, what you have attempted.

You know that any line passing through (1,0) must be of the form y= m(x-1).
What are the conditions on m so that it is tangent to both curves (at some points on those curves)?