Difference between d²y/d²x and d²y/dx²

[rohanprabhu]

Considering a parametric function:

<br /> y = sin(\theta)<br />
<br /> x = cos(\theta)<br />

I want to find it's second order derivative.

<br /> \frac{dy}{d\theta} = cos(\theta) <br />

<br /> \frac{dx}{d\theta} = -sin(\theta)<br />

Therefore, I can say that:

<br /> \frac{dy}{dx} = -cot(\theta)<br />

Also,

<br /> \frac{d^2y}{d\theta^2} = -sin(\theta)<br />

<br /> \frac{d^2x}{d\theta^2} = -cos(\theta)<br />

Hence,

<br /> \frac{d^2y}{d^2x} = tan(\theta)<br />

But, If i need the second order derivate, what i actually need is:

<br /> \frac{d^2y}{dx^2} = \frac{d}{d\theta}(\frac{dy}{dx})\times(\frac{d\theta}{dx})<br />

which turns out to be:

<br /> \frac{d^2y}{dx^2} = -cosec^3(\theta)<br />

So, basically, what is the difference between d²y/d²x and d²y/dx². And how can we arrive at the formula for d²y/dx²? I mean, what is the mathematical significance of that? And how to do it for higher order derivatives, let's say 3 or 4 for example.

Also, does something like d²y/d²x even exist? what does it mean?

 

[D H]

Also, does something like d²y/d²x even exist?

No, it does not exist. We specifically use the notation d²y/dx² to indicate that something more than mere division (as you did in the original post) is taking place.

 

[Gokul43201]

But, If i need the second order derivate, what i actually need is:

<br /> \frac{d^2y}{dx^2} = \frac{d}{d\theta}(\frac{dy}{dx})\times(\frac{d\theta}{dx})<br />

which turns out to be:

<br /> \frac{d^2y}{dx^2} = cosec(\theta)<br />

Are you sure that's right? I think I get a different answer using y =\sqrt{1-x^2}...

 

[jostpuur]

I'll explain two things. First how to do this

Considering a parametric function:

<br /> y = sin(\theta)<br />
<br /> x = cos(\theta)<br />

I want to find it's second order derivative.

<br /> \frac{dy}{d\theta} = cos(\theta) <br />

<br /> \frac{dx}{d\theta} = -sin(\theta)<br />

Therefore, I can say that:

<br /> \frac{dy}{dx} = -cot(\theta)<br />

more rigorously, and then why I think that this

Also,

<br /> \frac{d^2y}{d\theta^2} = -sin(\theta)<br />

<br /> \frac{d^2x}{d\theta^2} = -cos(\theta)<br />

Hence,

<br /> \frac{d^2y}{d^2x} = tan(\theta)<br />

But, If i need the second order derivate, what i actually need is:

<br /> \frac{d^2y}{dx^2} = \frac{d}{d\theta}(\frac{dy}{dx})\times(\frac{d\theta}{dx})<br />

which turns out to be:

<br /> \frac{d^2y}{dx^2} = cosec(\theta)<br />

is wrong.

If we first have functions

<br /> \theta\mapsto y(\theta)<br />
<br /> \theta\mapsto x(\theta)<br />

and you want to reparametrize y as a function of x, to get some function x\mapsto y, strictly speaking this is going to be different function y, so it would be clearer to give it a different notation. I'll put \tilde{y}. We must assume that we can invert the mapping \theta\mapsto x(\theta) to a mapping x\mapsto \theta(x), and then y and y-tilde are related by

<br /> \tilde{y}(x) = y(\theta(x))<br />

The derivative is then

<br /> \frac{d\tilde{y}(x)}{dx} = D_x y(\theta(x)) = \frac{dy(\theta(x))}{d\theta} \frac{d\theta(x)}{dx},<br />

or

<br /> \frac{dy}{d\theta}\frac{d\theta}{dx}<br />

for short, without parameters. In your example we must assume (for example) \theta &gt; 0 so that \theta(x)=\cos^{-1}(x) exists. Then we can substitute

<br /> \frac{dy}{d\theta} = \cos\theta<br />

<br /> \frac{d\theta}{dx} = -\frac{1}{\sqrt{1-x^2}} = -\frac{1}{\sin\theta}<br />

and get

<br /> \frac{d\tilde{y}}{dx} = -\frac{\cos\theta}{\sin\theta} = -\cot\theta<br />

You got this same result more easily, but check what the second derivative turns out to be.

<br /> \frac{d^2\tilde{y}(x)}{dx^2} = D_x\Big(\frac{dy(\theta(x))}{d\theta} \frac{d\theta(x)}{dx}\Big) = \Big(D_x \frac{dy(\theta(x))}{d\theta}\Big) \frac{d\theta(x)}{dx}\; + \;\frac{dy(\theta(x))}{d\theta}\Big( D_x \frac{d\theta(x)}{dx}\Big)<br />

<br /> = \Big(\frac{d^2 y(\theta(x))}{d\theta^2} \frac{d\theta(x)}{dx}\Big) \frac{d\theta(x)}{dx}\; +\; \frac{dy(\theta(x))}{d\theta} \frac{d^2\theta(x)}{dx^2} = \frac{d^2 y(\theta(x))}{d\theta^2} \Big(\frac{d\theta(x)}{dx}\Big)^2\; +\; \frac{dy(\theta(x))}{d\theta} \frac{d^2\theta(x)}{dx^2}<br />

or

<br /> \frac{d^2y}{d\theta^2} \Big(\frac{d\theta}{dx}\Big)^2 + \frac{dy}{d\theta} \frac{d^2\theta}{dx^2}<br />

for short, without parameters. By substituting

<br /> \frac{d^2y}{d\theta} = -\sin\theta<br />

<br /> \frac{d^2\theta}{dx^2} = -\frac{x}{(1-x^2)^{3/2}} = -\frac{\cos\theta}{\sin^3\theta}<br />

and some previous expressions, we get

<br /> \frac{d^2\tilde{y}}{dx^2} = -\sin\theta \frac{1}{\sin^2\theta}\; +\; \cos\theta\frac{-\cos\theta}{\sin^3\theta} = -\frac{1}{\sin^3\theta} = -\textrm{cosec}^3\theta<br />

 

[jostpuur]

Are you sure that's right? I think I get a different answer using y =\sqrt{1-x^2}...

Hehe. Or alternatively one can calculate

<br /> D^2_x \sqrt{1-x^2} = D_x \frac{-x}{\sqrt{1-x^2}} = \cdots = -\frac{1}{(1-x^2)^{3/2}}<br />



well, rohanprabhu, I hope you find the equation

<br /> \frac{d^2y}{dx^2} = \frac{d^2y}{d\theta^2} \Big(\frac{d\theta}{dx}\Big)^2 + \frac{dy}{d\theta} \frac{d^2\theta}{dx^2}<br />

interesting, at least. It could be useful if the mappings \theta\mapsto y(\theta) and \theta\mapsto x(\theta) were something different.

 

[rohanprabhu]

Are you sure that's right? I think I get a different answer using y =\sqrt{1-x^2}...

i'm not sure that this is right or not.. but I've been using this formula for my textbook questions.. and it sort of works. But since my textbooks doesn't really have advanced problems to deal with, i am not sure how applicable this formula is..

EDIT1:

no way.. i am totally wrong in this one. I tried doing using

<br /> y = \sqrt{1 - x^2}<br />

and I'm getting a totally different answer. I need to refer some valid source first..

EDIT2:

in the second method, I multiplied the values wrongly. I am not getting cosec(θ), but cosec³θ . But still, I'm losing a (-) sign somewhere.

@jostpuur: the post u made was p.e.r.f.e.c.t. Beautiful explanation.. thanks a lot.. now i truly understand it. and also.. ur formula was perfect...

P.S: you seriously typed all that in LaTeX?
P.S2: Could u tell me what exactly θ → x(θ) means.. does it mean something like, for a set of values of 'θ', we have a set of values of 'x' or that the argument of the function y-tilde varies as a function of 'θ'.

 

[jostpuur]

I later realized there's a chance, that I just didn't understand what

<br /> \frac{d^2y}{dx^2} = \frac{d}{d\theta}(\frac{dy}{dx})\times(\frac{d\theta}{dx})<br />

was supposed to mean, and that you had done a calculation mistake after this. Could you explain what this is?

P.S2: Could u tell me what exactly θ → x(θ) means.. does it mean something like, for a set of values of 'θ', we have a set of values of 'x' or that the argument of the function y-tilde varies as a function of 'θ'.

For example, if you want to define a mapping f:\mathbb{R}\to\mathbb{R}, f(x)=x^2, you could alternatively do it by talking about mapping \mathbb{R}\to\mathbb{R}, x\mapsto x^2. The arrow used to define how each member gets mapped is \mapsto, which is different from the usual one.

 

[jostpuur]

P.S: you seriously typed all that in LaTeX?

This is no-life.

 

[rohanprabhu]

I later realized there's a chance, that I just didn't understand what

\frac{d^2y}{dx^2} = \frac{d}{d\theta}(\frac{dy}{dx})\times(\frac{d\theta}{dx})

was supposed to mean, and that you had done a calculation mistake after this. Could you explain what this is?

This was a formula taught to us for the 2nd order derivative of a parametric function.

This is no-life.

totally ;)

 

[rohanprabhu]

Eureka.. the formula you provided and the one taught to me by my teacher is the same.. I spent some time over this, and arrived at this proof [finally :D]..

I have attached a screenshot of the proof, as I did the typeset in Mathematica [typing that much LaTeX would have been a pain...]