Differentiation (Show That Question)

**Simplicity** · May 4th 2008, 01:49 AM

Q: Given that $y = \mathrm{arcsin}\left( \frac{x}{a} \right)$ , where $a$ is a constant, show that $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - x \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^3 = 0$ .

My method:
$y = \mathrm{arcsin}\left( \frac{x}{a} \right)$
$\sin y = \frac{1}{a} . x$
$\cos y \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{a}$
$- \sin y \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2 + \cos y \left( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \right) = 0$

But that isn't what they require. What do I do?

Thanks in advance for the help.

**Moo** · May 4th 2008, 02:18 AM

Yop,

04/05/08: Today, I have made many threads for question. Thanks for all the help!

I've noticed that

Well, keep $y=\arcsin \frac xa$

$(\arcsin x)'=\frac{1}{\sqrt{1-x^2}}$

Hence, $\frac{dy}{dx}=\dots$

From here, you can get $\frac{d^2y}{dx^2}$ and $\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^3$

**Jhevon** · May 4th 2008, 02:22 AM

Originally Posted by Moo

Yop,

I've noticed that

Well, keep $y=\arcsin \frac xa$

$(\arcsin x)'=\frac{1}{\sqrt{1-x^2}}$

Hence, $\frac{dy}{dx}=\dots$

From here, you can get $\frac{d^2y}{dx^2}$ and $\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^3$

in other words. they want you to find dy/dx and then (d^2 x)/(dx^2) and plug it into the left hand side of the equation given, and show that you actually get 0

**Simplicity** · May 7th 2008, 11:37 AM

$y = \mathrm{arcsin}\left( \frac{x}{a} \right)$

$\therefore \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt {a^2 - x^2}}$

But where would I go from here? Finding the second derivative may be difficult.

**TheEmptySet** · May 7th 2008, 11:50 AM

Originally Posted by Air

$y = \mathrm{arcsin}\left( \frac{x}{a} \right)$

$\therefore \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt {a^ - x^2}}$

But where would I go from here? Finding the second derivative may be difficult.

$\frac{1}{\sqrt{a^2-x^2}}=(a^2-x^2)^{-1/2}$

taking the derivative from hear is simple....

$\frac{-1}{2}(a^2-x^2)^{-3/2}(-2x)=\frac{x}{(a^2-x^2)^{3/2}}$

Yeah!

**Chris L T521** · May 7th 2008, 11:54 AM

Originally Posted by Air

$y = \mathrm{arcsin}\left( \frac{x}{a} \right)$

$\therefore \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt {a^2 - x^2}}$

But where would I go from here? Finding the second derivative may be difficult.

There are two different ways of going through this. You can use the power rule or quotient rule.

Power Rule:

$\frac{1}{\sqrt{a^2-x^2}}=(a^2-x^2)^{-1/2}$

$\frac{d^2y}{dx^2}=-\frac{1}{2}(a^2-x^2)^{-3/2}(-2x)=\frac{x}{(a^2-x^2)^{3/2}}$

Quotient Rule:

$\frac{d^2y}{dx^2}=\frac{(a^2-x^2)^{1/2}(0)-(1)(\frac{1}{2}(a^2-x^2)^{-1/2}(-2x))}{a^2-x^2}=\frac{x}{(a^2-x^2)^{3/2}}$

Hope this helps out!!!

**bobak** · May 7th 2008, 01:01 PM

Originally Posted by Chris L T521

There are two different ways of going through this. You can use the power rule or quotient rule.

Power Rule:

$\frac{1}{\sqrt{a^2-x^2}}=(a^2-x^2)^{-1/2}$

$\frac{d^2y}{dx^2}=-\frac{1}{2}(a^2-x^2)^{-3/2}(-2x)=\frac{x}{(a^2-x^2)^{3/2}}$

Quotient Rule:

$\frac{d^2y}{dx^2}=\frac{(a^2-x^2)^{1/2}(0)-(1)(\frac{1}{2}(a^2-x^2)^{-1/2}(-2x))}{a^2-x^2}=\frac{x}{(a^2-x^2)^{3/2}}$

Hope this helps out!!!

I believe you meant to say the chain rule. and using the quotient rule for this question is long and pointless

This is slightly off topic, but seeing as Airs question has already been answered I doubt anyone will mind.

Just have a comment on your signature, you do realise that $i^i$ is multi-valued, I'll give an alternative proof of what you have written in your signature.

you can write $i$ as $e^{\frac{1+4n}{2} \pi i}$ for any integer n.

so $i^i$ becomes $(e^{\frac{1+4n}{2} \pi i})^i$

$\rightarrow e^{- \frac{1+4n}{2} \pi }$

Bobak

**Chris L T521** · May 7th 2008, 01:10 PM

Originally Posted by bobak

I believe you meant to say the chain rule. and using the quotient rule for this question is long and pointless

This is slightly off topic, but seeing as Airs question has already been answered I doubt anyone will mind.

Just have a comment on your signature, you do realise that $i^i$ is multi-valued, I'll give an alternative proof of what you have written in your signature.

you can write $i$ as $e^{\frac{1+4n}{2} \pi i}$ for any integer n.

so $i^i$ becomes $(e^{\frac{1+4n}{2} \pi i})^i$

$\rightarrow e^{- \frac{1+4n}{2} \pi }$

Bobak

I agree that the quotient rule is long and pointless, but it may be good to know that it can be done various ways. However, I did mean Power Rule. When we differentiate it, we end up using the chain rule any, for both the power rule and quotient rule.

Thanks for the input on the signature. Your adjustments to it somewhat makes sense.

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