# Homework Help: Another implicit differentiation Problem

1. Nov 4, 2005

### 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?

Last edited: Nov 4, 2005
2. Nov 4, 2005

### Fermat

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

3. Nov 4, 2005

### 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.

Last edited: Nov 4, 2005
4. Nov 4, 2005

### Fermat

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

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

5. Nov 4, 2005

### Fermat

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

6. Nov 4, 2005

### GregA

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)

7. Nov 4, 2005

### Fermat

Well, if

(y')² = 1 + 3sec²y

then you should get,

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

Edit: forgot to add the y' bit.

Last edited: Nov 4, 2005
8. Nov 4, 2005

### 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

Last edited: Nov 4, 2005
9. Nov 4, 2005

### Fermat

Yes.

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

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

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.

You said,

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 ?

Last edited: Nov 4, 2005
10. Nov 5, 2005

### 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.