Tricky (for me anyway) Differentiation.

**craigmain** · February 8th 2009, 01:33 AM

Hi

I am trying to differentiate.
Many attempts at trying, and I cannot get to the answer. I am doing something wrong. Just having the answer is not helping. I have tried using the product and quotient rule(s) but am probably not simplifying correctly. If someone can show the steps that would really be great.

$\frac{x^2-1}{\sqrt{x^2+1}}$

Thanks
Regards
Craig.

**mr fantastic** · February 8th 2009, 01:40 AM

Originally Posted by craigmain

Hi

I am trying to differentiate.
Many attempts at trying, and I cannot get to the answer. I am doing something wrong. Just having the answer is not helping. I have tried using the product and quotient rule(s) but am probably not simplifying correctly. If someone can show the steps that would really be great.

$\frac{x^2-1}{\sqrt{x^2+1}}$

Thanks
Regards
Craig.

Use the quotient rule. To get the derivative of the denominator, use the chain rule.

If you post your work it will be easier to know where you're going wrong.

**craigmain** · February 8th 2009, 03:15 AM

Here are some steps.

Here is the start.
$ \frac{2x.\sqrt{x^2+1}+\frac{x.\sqrt{x^2+1}}{(x^2+1 )^2}.(x^2-1)}{(x^2+1)} $

Then
$ \frac{(x.\sqrt{x^2+1}).(2(x^2+1)^2+(x^2-1))}{(x^2+1)^3} $

But something is wrong. I need to get to:

$ \frac{x.(x^2+3).\sqrt{x^2+1}}{(x^2+1)^2} $

I cannot see how simplifying will get the terms I need to cancel.

Cannot seem to get there. I have pages of working, and I am doing something stupid.

**Moo** · February 8th 2009, 03:41 AM

Hello,

Originally Posted by craigmain

Here are some steps.

Here is the start.
$ \frac{2x.\sqrt{x^2+1}+\frac{x.\sqrt{x^2+1}}{(x^2+1 )^2}.(x^2-1)}{(x^2+1)} $

Then
$ \frac{(x.\sqrt{x^2+1}).(2(x^2+1)^2+(x^2-1))}{(x^2+1)^3} $

But something is wrong. I need to get to:

$ \frac{x.(x^2+3).\sqrt{x^2+1}}{(x^2+1)^2} $

I cannot see how simplifying will get the terms I need to cancel.

Cannot seem to get there. I have pages of working, and I am doing something stupid.

You went wrong in the differentiation (1st step)

Quotient rule says :
$\left[\frac uv\right]'=\frac{u'v-uv'}{v^2}$

But what you did is $\frac{u'v-u \left(\frac 1v\right)'}{v^2}$

Because $-\frac{x\sqrt{x^2+1}}{(x^2+1)^2}$ is the derivative of $\frac{1}{\sqrt{x^2+1}}$ , not the derivative of $\sqrt{x^2+1}$
Just a careless mistake, I guess ^^'

**ADARSH** · February 8th 2009, 03:42 AM

Dont try to jump over the steps

I'll give you an example to differentiate

$ \frac{x-1}{\sqrt{x}} $
After using quotient rule the differntiation becomes

$ \frac{\sqrt{x} \frac{d(x-1)}{dx} - (x-1)\frac{d(\sqrt{x})}{dx} }{\sqrt{x}^2} $

Now try using it in your case and watch the use of quotient rule

**Moo** · February 8th 2009, 03:44 AM

Originally Posted by ADARSH

Dont try to jump over the steps

I'll give you an example to differentiate

$ \frac{x-1}{\sqrt{x}} $
After using quotient rule the differntiation becomes

$ \frac{\sqrt{x} \frac{d(x-1)}{dx} - (x-1)\frac{d(\sqrt{x})}{dx} }{\sqrt{x}^2} $

Now try using it in your case and watch the use of quotient rule

He indeed jumped some steps, but he did everything perfectly

(or maybe he did more on his paper)
It's just that he didn't apply the formula correctly :P

**ADARSH** · February 8th 2009, 04:04 AM

Originally Posted by craigmain

Here are some steps.

Here is the start.
$ \frac{2x.\sqrt{x^2+1}+\frac{x.\sqrt{x^2+1}}{(x^2+1 )^2}.(x^2-1)}{(x^2+1)} $

Maybe you are correct (And I I know it will be the case

)
But
I don't think differentiation of denominator in first case is correct

**HallsofIvy** · February 8th 2009, 06:48 AM

You might find it simpler to write you function as $(x^2- 1)(x^2+ 1)^{-\frac{1}{2}}$ and use the product rule.

**Krizalid** · February 8th 2009, 07:00 AM

$f(x)=\frac{{{x}^{2}}-1}{\sqrt{{{x}^{2}}+1}}=\frac{{{x}^{2}}+1-2}{\sqrt{{{x}^{2}}+1}}=\sqrt{{{x}^{2}}+1}-2{{\left( {{x}^{2}}+1 \right)}^{-\frac{1}{2}}},$ so

$f'(x)=\frac{x}{\sqrt{{{x}^{2}}+1}}+\frac{2x}{\sqrt {{{\left( {{x}^{2}}+1 \right)}^{3}}}}.$

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