Can anyone do this? I tried it so many time and keep getting the wrong answer, It's finding the derivative **using definition of derivative



This post was edited by Illumini on Feb 6 2014 04:29pm

Here is your setup:



does that agree with what you were doing? If so I'll go further.

the latex code for the above (in case someone wants to go with it)

f ' (x) = limit (as h -> 0) [ f(x+h) - f(x) ] / h

Let D = [ f(x+h) - f(x) ] / h

D = [ 3(x+h) / sqrt(x+h+4) - 3x / sqrt(x+4) ] / h

form one fraction :

D = [ 3(x+h).sqrt(x+4) -3x.sqrt(x+h+4) ] / [ h.sqrt(x+h+4).sqrt(x+4) ]

Expand a little in the numerator, then factorize by 3x :

D = [ 3x.(sqrt(x+4) - sqrt(x+h+4)) + 3h.sqrt(x+4) ] / [ h.sqrt(x+h+4).sqrt(x+4) ]

Separate into 2 fractions :

D = [ 3x.(sqrt(x+4) - sqrt(x+h+4)) ] / [ h.sqrt(x+h+4).sqrt(x+4) ] + 3 / sqrt(x+h+4)

Only the limit of the first fraction is not obvious. Like most of square roots issues, the trick is to use the conjugate :

sqrt(a) - sqrt(b) = ( a - b ) / ( sqrt(a) + sqrt(b) )

Hence :

D = - 3x / [ ( sqrt(x+4) + sqrt(x+h+4) ).sqrt(x+h+4).sqrt(x+4) ] + 3 / sqrt(x+h+4)

and now it's easy to find the limit as h -> zero :

f ' (x) = - 3x / [ 2.sqrt(x+4) ^3] + 3 / sqrt(x+4)

f ' (x) = ( -3x + 6(x+4) ) / [ 2.sqrt(x+4) ^3]

f ' (x) = (3x+24) / [ 2.sqrt(x+4) ^3]

Yes it does, could you finish with that program because

this is pretty hard to follow

if you could write this out on paper or a program, or explain it in more steps it would be beyond appreciated

ok give me a few and i'll whip up something

My solution differs slightly from feanur's. His is correct, mine is wrong. I clearly have a sign problem, and I'm too tired to track it down

Track down the mistake, be good for you!

oh i see it now, its between step 2 and 3... that trailing fraction in line 3 should be added to the first fraction, not subtracted!

Big thanks for giving a clean answer, I had not enough time (and energy) yesterday night to do so !