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WE KNOW 3^2=3+3+3 5^2=5+5....5 TIMES SO x^2=x+x+x.....xtimes differentiating both sides d(x^2)/dx=d(x+x+....x times)/dx =>2x=1+1+1....xtimes =>2x=x so, 2=1 adding 3 on both sides we get 5=4 !! BUT WE KNOW d(x^2)/dx=2x but here we get x...how ?? whats the mistake here plz tell....

WE KNOW


3^2=3+3+3


5^2=5+5....5 TIMES


SO x^2=x+x+x.....xtimes


differentiating both sides


d(x^2)/dx=d(x+x+....x times)/dx


=>2x=1+1+1....xtimes


=>2x=x


so, 2=1


adding 3 on both sides


we get


5=4  !!


BUT WE KNOW


d(x^2)/dx=2x


but here we get x...how ??


whats the mistake here plz tell....

Grade:

3 Answers

abhishek f
14 Points
9 years ago

there is a problem in diff. d expression.

when u write x+x+x+x }x times , u put a constraint on x dat it can take only natural number values.

Therefore its grapgh will be points on integral values(+ integers) and not a continuos function. It will be discrete and hence non differentiable.

 

x=1 y=1

x=2 y=4

but x=0.4 is not in domain

Ranjita yadav
35 Points
9 years ago

In differential , formula is d(xn)/dx = n * x n-1

hence , d(x2)/dx = 2*x2-1 =2x

Ashwin Muralidharan IIT Madras
290 Points
9 years ago

Hi Shubam,

 

x^2 = x + x + x +..... xtimes, is valid only when x is an integer.

So this equation is valid only for integral values of x.

So x cannot take non-integral values.

If you plot the graph, it is discontinuous for the entire range of real numbers.

As it is not continuous, you cannot differentiate it (that is where the mistake is).

 

Regards,

Ashwin (IIT Madras).

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