Question

Differentiate the below equation :-

Answer 1

\[\huge \bf \frac{d}{dx}(\frac{2}{sinx})\]

Answer 2

@ranga

Answer 3

You'll need to use the chain rule. Or the quotient rule if you feel more comfortable with that.

Answer 4

@primeralph @zzr0ck3r

Answer 5

How to use chain rule in this equation ?

Answer 6

\[\frac{ dy }{ dx }=\frac{ dy }{ du }\frac{ du }{ dx }\] You need to make a u-substitution which an appropriate choice for u.

Answer 7

With not which*

Answer 8

@hartnn @ganeshie8

Answer 9

Using the chain rule we got:
\[\frac{ d }{ dx }(\frac{ 2 }{ sinx })=2\frac{ d }{ dx }(\frac{ 1 }{ sinx })=2\frac{ (-\sin(x))' }{ \sin^{2}(x) }\]

Answer 10

Do you understand @needforspeed

Answer 11

no....

Answer 12

can you tell me how to use chain rule and where?

Answer 13

Look at what I've done:
1.First I've bringed the constant in front of the diferential "2"(You can do these all the time it's a property of differentials)
2.Then I've differentiate 1/sin(x) with the formula:
\[\frac{ d }{ dx }(\frac{ 1 }{ u(x) })=-\frac{ u(x)' }{ u^{2}(x) }\]

Answer 14

If it's something unclear ask me please

Answer 15

if we apply quotient rule,then

Answer 16

You can apply qutient rule as well instead knowing the formula
The result is the same

Answer 17

see my way of solution and tell me whether i am wrong or right,ok?

Answer 18

okay

Answer 19

\[\large \bf \frac{d}{dx}\frac{2}{sinx}\]
\[\large \bf \frac{sinx(0)-2(cosx)}{(sinx)^2}\]
\[\large \bf \frac{-2cosx}{\sin^2x}\]

Answer 20

Perfect

Answer 21

this is the answer,right

Answer 22

YES

Answer 23

@ifrimpanainte can you please tell me how to do with chain rule?

Answer 24

i don't practise problems regarding with denominator as a trig.function

Answer 25

so can you please explain me???

Answer 26

OMG!!! g2g..come back soon..........

so can you write your solution with explanation applying chain rule and how to apply that rule...

Answer 27

i hope you write your solution very nicely..

Answer 28

@ifrimpanainte

Answer 29

I've applied the chain rule in my first post here

Answer 30

The standard formula is:
\[\frac{ d }{ dx }(\frac{ 1 }{ x })=-\frac{ 1 }{ x^2 }\]
Now when you substitute the x with u(x) you are using the chain rule:
You replace x with u(x) and multiply with (u(x))'
\[\frac{ d }{ dx }(\frac{ 1 }{ u(x) })=-\frac{ u(x)' }{ u^{2}(x) }\]