Simple Calculus Question

**evanator** · July 7th 2010, 03:44 AM

Hi there,

There is no forum for basic calculus, so I will have to post in the university-level forum. I have been trying to refresh my maths for the last few months (I had forgotten everything, right down to how to do long division.) and I feel as if I have been making good progress.

I have a question about a very simple differentiation exercise:

Using the formula

$f'(x) = \displaystyle\lim_{\delta x \to 0} \frac{f(x + \delta x) -f(x)}{\delta x}$

find the derivative of $\frac{1}{x^2}$

Here is my attempt at a solution:

$f(x + \delta x) = \frac{1}{(x + \delta x)^2}$

$f(x + \delta x) - f(x) = \frac{1}{x^2 + 2x \delta x + (\delta x)^2} - \frac{1}{x^2}$

$= \frac{x^2 - (x^2 + 2x \delta x + (\delta x)^2)}{(x^2 + 2x \delta x + (\delta x)^2)x^2}$

$= \frac{\delta x(- \delta x -2x)}{x^4 + 2x^3 \delta x + x^2(\delta x)^2}$

$\frac{f(x + \delta x) - f(x)}{\delta x} = \frac{\delta x(- \delta x - 2x)}{x^4 + 2x^3 \delta x + x^2(\delta x)^2} \times \frac{1}{\delta x}$

$= \frac{-2x - \delta x}{x^4 + 2x^3 \delta x + x^2(\delta x)^2}$

$f'(x) = \displaystyle\lim_{\delta x \to 0} \frac{-2x - \delta x}{x^4 + 2x^3 \delta x + x^2(\delta x)^2}$

$= \frac{-2x}{x^4}$

$= \frac{-2}{x^3}$

It would be great if someone would tell me where I am going wrong.

Regards,

Evanator

**ramiee2010** · July 7th 2010, 04:18 AM

Originally Posted by evanator

Hi there,

There is no forum for basic calculus, so I will have to post in the university-level forum. I have been trying to refresh my maths for the last few months (I had forgotten everything, right down to how to do long division.) and I feel as if I have been making good progress.

I have a question about a very simple differentiation exercise:

Using the formula

$f'(x) = \displaystyle\lim_{\delta x \to 0} \frac{f(x + \delta x) -f(x)}{\delta x}$

find the derivative of $\frac{1}{x^2}$

Here is my attempt at a solution:

$f(x + \delta x) = \frac{1}{(x + \delta x)^2}$

$f(x + \delta x) - f(x) = \frac{1}{x^2 + 2 \delta x + (\delta x)^2} - \frac{1}{x^2}$

..............

you forgot x in second step ,
$f(x + \delta x) - f(x) = \frac{1}{x^2 + 2x \delta x + (\delta x)^2} - \frac{1}{x^2}$

now continue....

**mr fantastic** · July 7th 2010, 05:01 AM

Originally Posted by evanator

Hi there,

There is no forum for basic calculus, so I will have to post in the university-level forum.

[snip]

Well done. Thankyou. Someone with some basic common sense. And you even showed your work.

**evanator** · July 7th 2010, 05:19 AM

I amended the first post. It all comes out now. Thanks for the help, ramiee2010. It is much appreciated.
What are you referring to, mr fantastic?

**Prove It** · July 7th 2010, 05:53 AM

Originally Posted by evanator

I amended the first post. It all comes out now. Thanks for the help, ramiee2010. It is much appreciated.
What are you referring to, mr fantastic?

He's saying a lot of people post calculus questions in the pre-calculus forum just because they happen to be in high school and not university.

And he's also giving you praise for showing your working, as a lot of people also just post a question and beg for others to do the work for them.

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