let-S-i-a-n-x-i-prove-or-disprove-that-d-dx-i-a-n-x-i-i-a-n-d-dx-x-i-i-a-n-x-i-

Question Number 3385 by Filup last updated on 12/Dec/15

let :

S = \underset{i = a}{\sum^{n}} x_{i}

prove or disprove that :

\frac{d}{d x} (\underset{i = a}{\sum^{n}} x_{i}) = \underset{i = a}{\sum^{n}} (\frac{d}{d x} {x_{i}})

= \underset{i = a}{\sum^{n}} x_{i}^{^{'}}

Commented by prakash jain last updated on 12/Dec/15

Some of two function which are not differentiable

at x = x_{0} may be differentiable at x = x_{0} .

Commented by Filup last updated on 12/Dec/15

i w a s m e a n t t o w r i t e t h a t

x_{i} i s t h e i^{th} t e r m i n a s e r i e s

Commented by prakash jain last updated on 12/Dec/15

It does make a difference . If x_{i} i^{t h} term in

series . It is a still function of x .

Commented by Filup last updated on 12/Dec/15

i see . hmm \dots so is there ever a time

such that the above is true ?

other than x_{i} = constant

Commented by prakash jain last updated on 12/Dec/15

It is generally true if each of x_{i} is differentiable .

I gave an exception rather than rule .

Commented by Filup last updated on 12/Dec/15

Ah I understand now, thanks!

Commented by prakash jain last updated on 12/Dec/15

\frac{d}{d x} (f (x) + g (x)) = \frac{d}{d x} f (x) + \frac{d}{d x} g (x)

w h a t y o u h a v e a b o v e i s a e s s e n t i a l l y s a m e

f o r m u l a .

Answered by prakash jain last updated on 12/Dec/15

x_{1} = {\begin{cases} 2 & x ⩾ 1 \\ 3 & x < 1 \end{cases}

x_{2} = {\begin{cases} - 2 & x ⩾ 1 \\ - 3 & x < 1 \end{cases}

\frac{d}{d x} (x_{1} + x_{2}) at x = 1 is not equal to \frac{d}{d x} x_{1} + \frac{d}{d x} x_{2}

at x = 1 .