I very confuse with the whole "derivative" concept.

Attachment 28487

Really appreciated if someone can tell me how to work this out and get the solution for this example.

thanks guys

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- May 29th 2013, 08:47 PMJamesT1st principle derivative
I very confuse with the whole "derivative" concept.

Attachment 28487

Really appreciated if someone can tell me how to work this out and get the solution for this example.

thanks guys - May 29th 2013, 11:31 PMchiroRe: 1st principle derivative
Hey JamesT.

Try just unpacking the definition first. As a starting hint consider that if f(x) = -1 / [x^2 - x] then f(x+h) = -1 / [(x+h)^2 - (x+h)]. If you let triangle_x = h, then the point is to cancel out the h on the denominator and evaluate the limit. - May 29th 2013, 11:38 PMMINOANMANRe: 1st principle derivative
James

It is not dfficult but has an ocean of tricky operations and fractions....try slowly slowly and do it...replace x as x+Δx in the first function and subtruct the initial function.do whatever can be done to simplify it and then divide by Δx .get the limits and find the final answer f'(x) =(2x-1)/x^2(x-1)^2 - May 30th 2013, 04:39 AMHallsofIvyRe: 1st principle derivative
$\displaystyle f(x)= \frac{-1}{x^2- x}$ so $\displaystyle f(x+ \Delta x)= \frac{-1}{(x+ \Delta x)^2- (x+ \Delta x)}$

Rather than multiplying that out it is simplest to factor: $\displaystyle (x+ \Delta x)^2- (x+ \Delta x)= (x+ \Delta x)(x+ \Delta x- 1)$

So $\displaystyle f(x+ \Delta x)- f(x)= \frac{-1}{(x+ \Delta x)(x+ \Delta x- 1)}- \frac{-1}{x(x- 1)}$

To subtraction we need to get the "common denominator" and, since there are no common factors, we can do that by multiplying both numerator and denominator of each by the denominator of the other:

$\displaystyle \frac{-x(x- 1)}{x(x- 1)(x+ \Delta x)(x+ \Delta x- 1)}- \frac{-(x+ \Delta x)(x+ \Delta x- 1)}{x(x- 1)(x+ \Delta x)(x+ \Delta x- 1)}$

Ignore the denominator for the moment and multiply out the numerators:

$\displaystyle -x^2- x+ x^2- 2x\Delta x+ (\Delta x)^2+ x- \Delta x= (\Delta x)^2- (2x- 1)\Delta x= \Delta x(\Delta x- 2x+ 1)$

Notice that every term that does not have $\displaystyle \Delta x$ in it cancelled. That**had**to happen because when $\displaystyle \Delta x= 0$ the two terms would be the same.

So far we have $\displaystyle f(x+ \Delta x)- f(x)= \frac{\Delta x(\Delta x- 2x+ 1)}{x(x- 1)(x+ \Delta x)(x+ \Delta x- 1)}$

where I have just included that common denominator from before.

Next we want $\displaystyle \frac{f(x+\Delta x)- f(x)}{\Delta x}$, the "difference quotient". But it is easy to see that the $\displaystyle \Delta x$ in the denominator will just cancel that $\displaystyle \Delta x$ I factored out of the numerator.

All that is left to do is to take the limit as $\displaystyle \Delta x$ goes to 0. But now that there is no $\displaystyle \Delta x$ in the denominator, we can just replace $\displaystyle \Delta x$ in the formula with 0:

$\displaystyle \frac{0- 2x+ 1}{x(x- 1)(x+ 0)(x+ 0- 1)}= \frac{-2x+ 1}{x(x-1)x(x-1)}= \frac{-2x+ 1}{x^2(x- 1)^2}$

(You will be happy to know that you will soon learn general formulas so you won't have to do this for every derivative!) - May 30th 2013, 10:39 PMJamesTRe: 1st principle derivative
Thank you so much guys!!!