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dse doubt

SoniaKapoor
413 posts
MA Economics
DSE
2014-16
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Re: dse doubt

Akshay Jain
584 posts
This post was updated on Jun 21, 2014; 8:20pm.
we can view 1st expression as an expression of left hand derivative and it is given that the function is differentiable in its domain so this must be true in general
the 2nd statement wil not hold valid in general.
Akshay Jain
Masters in Economics
Delhi School of Economics
2013-15
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Re: dse doubt

mrittik
192 posts
In reply to this post by SoniaKapoor
we can do it by 1st principal of limit ie f'(x)=lim h-0  f(x)-f(x-h)/x-(x-h)=f(x)-f(x-h)/h

for the second one f'(x)=lim h-0 f(x+2h)-f(x+h)/((x+2h-(x+h))=f(x)-f(x-h)/h

understood? @Akshay-boss plz check it & comment
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Re: dse doubt

mrittik
192 posts
we can do it by 1st principal of limit ie f'(x)=lim h-0  f(x)-f(x-h)/x-(x-h)=f(x)-f(x-h)/h

for the second one f'(x)=lim h-0 f(x+2h)-f(x+h)/((x+2h-(x+h))= f(x+2h)-f(x+h)/h
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Re: dse doubt

Akshay Jain
584 posts
yes mrittik i think ur way is correct.....great work...
Akshay Jain
Masters in Economics
Delhi School of Economics
2013-15
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Re: dse doubt

mrittik
192 posts
thanks Akshay....
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Re: dse doubt

SoniaKapoor
413 posts
Thanx both of u
MA Economics
DSE
2014-16