Simple Calculus Question

Mar 2010
24
1
Ireland
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

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

find the derivative of \(\displaystyle \frac{1}{x^2}\)

Here is my attempt at a solution:

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

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

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

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

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

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

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

\(\displaystyle = \frac{-2x}{x^4}\)

\(\displaystyle = \frac{-2}{x^3}\)

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

Regards,

Evanator
 
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May 2009
153
70
New Delhi
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

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

find the derivative of \(\displaystyle \frac{1}{x^2}\)

Here is my attempt at a solution:

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

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

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

now continue....
 
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mr fantastic

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Dec 2007
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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. (Clapping)
 
Mar 2010
24
1
Ireland
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?
 

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Aug 2008
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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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