Differential Equations...

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September 12th 2008, 07:34 PM
yeloc

Differential Equations...

Use the definition of the derivative to find f' (x).

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

f'(x)= ${\lim_{\Delta x \rightarrow 0}} \frac{\frac {1}{(x+ \Delta x)^2} - \frac {1}{x^2}} {\Delta x}$

Should I get a common denominator in the top of the problem?
September 12th 2008, 07:44 PM
Shyam

Quote:

Originally Posted by yeloc

Use the definition of the derivative to find f' (x).

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

f'(x)= ${\lim_{\Delta x \rightarrow 0}} \frac{\frac {1}{(x+ \Delta x)^2} - \frac {1}{x^2}} {\Delta x}$

Should I get a common denominator in the top of the problem?

yes. you should get a common denominator in the top, and simplify the top of numerator.
September 12th 2008, 07:56 PM
yeloc

I got a common denominator of $x^4+2x^3 \Delta x+x^2 \Delta x^2$

So would the final answer be $\frac {-2x} {x^4}$ ?
September 12th 2008, 08:23 PM
mr fantastic

Quote:

Originally Posted by yeloc

I got a common denominator of $x^4+2x^3 \Delta x+x^2 \Delta x^2$

So would the final answer be $\frac {-2x} {x^4}$ ?

The answer is correct but can be simplified.
September 12th 2008, 08:29 PM
yeloc

$\frac {-2} {x^3}$ I knew I missed something. Thanks!(Clapping)

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