Another implicit differentiation Problem

[GregA]

The question I'm having trouble with is as follows:
Given that siny = 2sinx show that:
a) (dy/dx)^2 = 1+3sec^2(y), by differentiating this equation with respect to x show that
b)d^2y/dx^2 = 3sec^2ytany and hence that
c) coty(d^2y/dx^2) - (dy/dx)^2 + 1 = 0
Part (c) is straight forward and after a fair bit of work I got (a)...part (b) however is a *big* problem for me.
my favourite method of trying to solve the problem (because this question doesn't come anywhere close to the 4 examples the book has shown up to now) is to firstly to find the square root of both sides of the equation to get back to dy/dx and then differentiate working on the principle that this is the square root of a quotient.

dy/dx = sqrt(1+ 3sec^(2)y)...
(y'') = 1/2(y')(6cosysiny/cos^4(y))(1 + 3sec^2(y))^(1/2)...
(y'') = (y')(3sec^ytany)(1 + 3sec^2)^(1/2)
I am stuck here with the bit that I want nested within rubbish
can someone please show me what I should have done to reach the correct answer?

 

[Fermat]

I found it easier to just differentiate (implicitly) twice.

 

[GregA]

Thanks for the reply fermat though I'm still having problems!
If I just differentiate twice the equation siny = 2sinx...
differentiating once...cosy(y') = 2cosx and therefore (y') = 2cosx/cosy
differentiating twice...(y')(-siny) + (y'')cosy = -2sinx
by substituting for (y') and using 2sinx = siny I get...
(y'')cosy = 2cosxsiny/cosy - siny...
(y'') = 2cosxsiny/cos^2(y) - tany...and I am getting nowhere

I can differentiate 1+3sec^2(y) neither once or twice implicitly because 1+3sec^2(y) is the square of (y') and so I would not get (y'') by doing this...that was my reason for taking the square root of both sides, and then differentiating

As with many of the questions I have dealt with before this, it feels like its a massive jump from what has been explained in the book...and studying in my spare time with no one in my circle of friends, family, or colleagues to help me I have to just *infer* certain rules that have not been covered or even touched upon...more often than not I can fill in the gaps but here I'm struggling.

 

[Fermat]

...
differentiating once...cosy(y') = 2cosx and therefore (y') = 2cosx/cosy
differentiating twice...(y')(-siny) + (y'')cosy = -2sinx
...

There's the error!
When you differentiate cosy(y'), you should get,

(-siny*.y')*y' + cosy*y''

 

[Fermat]

...
I can differentiate 1+3sec^2(y) neither once or twice implicitly because 1+3sec^2(y) is the square of (y') and so I would not get (y'') by doing this...that was my reason for taking the square root of both sides, and then differentiating
...

I'm not too sure I follow your reasoning here, but if you differentiate (y')² you get,
2y'*y''
dunno if that would help though

 

[GregA]

I'm not too sure I follow your reasoning here, but if you differentiate (y')² you get,
2y'*y''
dunno if that would help though

it does help (because I overlooked that!), though I still can't see a solution to my problem with it because now I get:
2(y')(y'') = (y')(3sec^ytany)(1 + 3sec^2)^(1/2)
2(y'') = (3sec^ytany)(1 + 3sec^2)^(1/2)

 

[Fermat]

Well, if

(y')² = 1 + 3sec²y

then you should get,

2y'.y'' = 6secy*secy.tany.y'

Edit: forgot to add the y' bit.

 

[GregA]

ah...I can see where you are coming from now but with reference to what you said about my reasoning before...isn't (y')^2 (y')(y')?...but you can express (y'') as (y')(y') also... but the two are not the same...and by differentiating (1 + sec^2y) I am finding the derivative of a squared derivative, not the second derivative...agh! either I am stupid...the book's author expects me to be at college whilst I use it...or both

p.s. noticed the y' bit but knew what you meant without it

 

[Fermat]

...isn't (y')^2 (y')(y')?...

Yes.

..but you can express (y'') as (y')(y') also...

Nope!
y'' is the 2nd derivative. (y')(y'), or (y')² is the square of the 1st derivative.

$$\frac{d^2y}{dx^2} \mbox{ is not = } \left(\frac{dy}{dx}\right)^2$$

...and by differentiating (1 + sec^2y) I am finding the derivative of a squared derivative, not the second derivative...

The derivative of a squared (1st) derivative will give you the 2nd derivative.
(y')² when differentiated gives 2y'*y'', which includes both the 1st derivative and the 2nd derivative.

ah...I can see where you are coming from now but with reference to what you said about my reasoning before...

You said,

I can differentiate 1+3sec^2(y) neither once or twice implicitly because 1+3sec^2(y) is the square of (y') and so I would not get (y'') by doing this...

and I didn't underestand why you wouldn't get y'' for the reason you gave, that 1+3sec²y is the square of y'.

Anyway, have you solved the problem now ?

 

[GregA]

The problem is solved now wasn't comfortable with (y')^2 because I had neither seen it or knew what it would yield if differentiated.
Thankyou for your help