Question

prove that :
if y=cosecx
y'=-cosecxcotx

Answer 1

@amrmagdy we know
\[cosec\ x=\frac{1}{\sin x}\]
\[y=cosec\ x=\frac{1}{\sin x}\]Do you know quotient rule of differentiation?

Answer 2

yes

Answer 3

Can you use it to prove?

Answer 4

1 /cosx ??

Answer 5

ohhhhhhhhhhhhhhhhhhhhh

Answer 6

you mean division rule of 1/sinx ??

Answer 7

yes

Answer 8

:D , good , can we prove it using limits ??

Answer 9

Yeah sure
let's do that
\[f(x)=cosec \ x\]
By the definition of the derivative
\[f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{x+h-x}\]
Do you get this?

Answer 10

man , you are just perfect . when respect is not enough:))

Answer 11

but will be lim h+x>>>x ??

Answer 12

But we aren't done yet ;)
\[\lim_{h\to 0}\frac{cosec (x+h)-\ cosec (x)}{h}\]
\[\lim_{h\to 0}\frac{\frac{1}{\sin (x+h)}-\frac{1}{\sin (x)}}{h}\]
\[\lim_{h\to 0}\frac{\frac{\sin x-\sin (x+h)}{\sin x\times \sin (x+h)}}{h}\]

Answer 13

right . i got it nw

Answer 14

Limit will be \(h\to 0\)

Answer 15

@amrmagdy could you do it from here or you want me to do?

Answer 16

no thanks i got it all .. how old are u sir ?? ur a teacher??

Answer 17

@amrmagdy Don't be formal, I'm a student like you :)

Answer 18

about 600 fans . 2000 medals .. 3000 answered question . and you are like me !!.. hw come ?? ^_^

Answer 19

It's because I'm here on Open Study for a while. You'd also get these soon enough :)

Answer 20

hope so :) .. snx mate