21Wen u say f(x) is "x" times x.....
U treat the bold "x" as a constant say k.....
ther4 f(x) = k(x)
ther4 f'(x) = k ....
but ur k =x so ur 2nd f(x) turns out 2 be x... [3]
Let f(x)=x2
but x2=x times x
=x+x+x+x+x+..... x times
=>f'(x)=1+1+1+1+1+....x times
=x
But dx2/dx=2x!
How?!
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14 Answers
21
1Close...
X2=x times x only for +ve integral values!
Thus f(x)=x times x is not a continuous function because it is not defined for all x.
Hence its derivative is meaningless.
Although tapanmast's approach is also interesting...
9i guess the ans may deal with the fact that the representation of x times x also depends on x
1the function x2 is continuous whereas x +x+ .... is not a continous function
33it was once in qod also, i think..
also 1/x+1/x+1/x.... x times =1
now differenciate it...
-1/x2-1/x2.... x times =0
-1/x=0
[4]
62guys.. it is bad.. specially some of the old ones..
we had done this on Question of the day once..
this question has already come up twice before for discussion :(
1the function should be continuos for it tobe differentiable which is not the case here.
1hey tapan,I think
x2=x.x is ok bcoz
f(x)=x2=x.x
=> f '(x)= d(x.x)/dx=x(dx/dx) + x(dx/dx)=2x
so there is no prob. in writing x2 as x.x.
Also, there is no such continuity-differentiability prob.
Error creeps in when you write x.x=x+x+x............x times
bcoz when we write nx=x+x+x+....... n times n is a const there4 n can't be replaced by x itself.