IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - Contradictory answers for the same question (on odd/even function)

Prakriteesh

Q: The derivative of an even function is an even function or an odd function?

The problem is that if we substitute f( -x) = f(x) in the d/dx operator we get it to be an even function but if we solve by first-principle we get that it is an odd function. Where am I mistaken?

vish0001

sorry not able to get u !!

Prakriteesh

Let f(x) be an even function. Then f(x) = f(-x) = g(x) (say)

then d/dx of f(x) = d/dx of g(x) ..........................(1)

and d/dx of f(-x) = d/dx of g(x) ..........................(2)

So, d/dx of f(x) = d/dx of f( -x)

So, the derivative of an even function is an even function.

But from first principle,

d/dx of f(x) =

_{[ h]}

_[0] {f(x+h) - f(x)}/h

d/dx of f( -x) = _{[ h][0]}{f(-x+h) - f( -x)}/h = _{[ h][0]} {f(-(x-h)) - f (-x)}/h

= _{[ h][0]} {f(x-h) - f(x)}/h since f is an even function.

= _{[ m][0]} {f(x+m) - f(x)}/(-m) Putting h =( -m)

=[ - _{[ m][0]} {f(x+m) - f(x)}/m] = -d/dx of f(x)

So, from this we get that the derivative of an even function is an odd function. Where is the mistake?

vish0001

yup i got ur mistake in the 1 st principal u have used , u have taken h= -m , ok m tends to 0 . but u forgot that the limit is always applicable with the quantity h always +ve !!!! m is definitely -ve

so d/dx of f(x) = f(x) - f(x-h) /h and not that u have written ! when u express it like this the minus minus will cancel and give the same result !!!

Prakriteesh

I knew that I was going to get that answer from someone, coz that is what I thought first (REALLY!). But I am afraid it is not correct. Recall how we show that a function is derivable or not by proving whether L.H.D. = R.H.D.

At the time of taking out L.H.D. we consider h to be negative number approaching 0.

So, L.H.D. = _{[ h][0^- ]} { function (x+h) - function (x)}/h. Thus, h need not always be positive.

Therefore, I again ask where is my mistake?

puneet

hii

well yaar derivative of even function f will be odd only ..

see u try to see wat property does f '(x) satisfy ... okay ??

now d/dx of f(x) = d/dx of f( -x) .. here u have done it wrong na ..

d/dx of f( -x) is not the derivative of f ..

do u get wat i am trying to say ..

we get f ' (x) = - f ' (-x)

so the function is odd ..

okay .. cheers

Prakriteesh

Puneet da, sorry but I didn't understand why "d/dx of f( -x) is not the derivative of f". Since f(x) = f(-x) so by putting both the functions under d/dx operators, aren't the differentiations same.

Prakriteesh

I think I got it! The problem is not due to the function but due to the variable w.r.t. which we are differentiating. Thus if f(x) = f(-x) then d/dx of f(x) is not equal to d/dx of f( -x), but d/dx of f(x) = d/d(-x) of f(-x) = {d/dx of f(-x)}X{d/d(-x) of x} = -d/dx of f(x). So, it is an odd function. Thanks.

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