Question

Solve it :-

Answer 1

\[\huge \bf \frac{d}{dx}(-u)\]

Answer 2

just differentiate it..

Answer 3

\[\huge \bf -\frac{du}{dx}\]

Answer 4

how?

Answer 5

please explain it

Answer 6

-u = -1 * u
d/dx(-u) = d/dx(-1 * u) = -1 * du/dx

Answer 7

\[\large \bf -1 \times \frac{d(u)}{dx}=-1 \times 1 \times u^{1-1}=-1\]

Answer 8

if 'u' is the function of x, then yes, -du/dx as you can throw a constant out of differentiation and -1 is a constant.
if u is not a function of x
then d/dx(-u) = 0
as 'u' will be treated as constant. and derivative of constant = 0

Answer 9

isn't this correct? @ranga

Answer 10

you're differentiating w.r.t \(\Huge x\)
and not 'u'
d/du (u) would be = 1

Answer 11

and d/dx (x) would be 1

but d/dx (*anything not a function of x*) = 0
as *anything not a function of x* would be treated as constant while differentiating w.r.t x

Answer 12

doubts ?

Answer 13

this is wrong or correct ? @hartnn

Answer 14

no, thats incorrect.

Answer 15

why?

Answer 16

d/dx(-u) = -du/dx
We don't know what u is.
If u is a constant or u is some other function not dependent on x then du/dx will be 0.
But if u is a function of x, we have to know what u is before we can actually differentiate u.
In either case you can leave it as -du/dx and when u is supplied we will know if it is 0 or we need to differentiate further. Until then the answer is just -du/dx.

Answer 17

because the variable you are differentiating, the 'u'
and the variable w.r.t which the differentiation is supposed to take place, the 'x'
are different variables.
you can apply differentiation rules only if you are differentiating the same variable, or the function of that variable.

Answer 18

can you give me some example to prove your statement ? ( i can't actually understand) @hartnn

Answer 19

@ranga if u is a function of x,then the answer would be -1.
Correct ?

Answer 20

Not in general. But if u = x then yes.
If u = x^2 then du/dx = 2x
if u = 15, then du/dx = 0
if u = y^3, then du/dx = 0.

Answer 21

\(\large \dfrac{d}{dx}f(x) = f'(x) \\~ \\ \large \dfrac{d}{du}f(x) = 0\)
because you are differentiating w.r.t u, but differentiating f(x)

\(\\ \large \dfrac{d}{dx}f(u) = 0\)

Answer 22

if i am correct,we don't know the function of x w.r.t u, thats why we left to differentiate it.

Answer 23

correct ? @ranga

Answer 24

d/du (u) = 1 * u^(1-1) = 1
d/dx (u) = du/dx

Answer 25

if you don't know whether u is the function of x or not,
then its better to keep it as -du/dx
as ranga said

Answer 26

oh!!!! thank you so much....guys

Answer 27

now i understood...

Answer 28

You are welcome. :)