# A level differentiation chain rule

idk how to get just cosecx i keep getting 1/2cosec1/2x

https://imgur.com/a/XV6QgqX
Original post by esha06
idk how to get just cosecx i keep getting 1/2cosec1/2x
https://imgur.com/a/XV6QgqX

You lose the square on sec^2. Also be clearer about whether its 1/2 or 1/(1/2)
(edited 2 months ago)
Original post by mqb2766
You lose the square on sec^2. Also be clearer about whether its 1/2 or 1/(1/2)

that makes my problem…worse 😭😭😭😭😭

https://imgur.com/a/3IFWidG
Original post by esha06
that makes my problem…worse 😭😭😭😭😭
https://imgur.com/a/3IFWidG

Looking at the answer, youd expect the denominator to be
2cos(x/2)sin(x/2)
so how is it (1/2)...?
Original post by mqb2766
Looking at the answer, youd expect the denominator to be
2cos(x/2)sin(x/2)
so how is it (1/2)...?
bc ur doing chain rule with tan1/2 x, how am i meant to get rid of the 1/2 x
Original post by esha06
bc ur doing chain rule with tan1/2 x, how am i meant to get rid of the 1/2 x

1/2 sec^2 = 1/2 * 1/cos^2 = 1/(2cos^2 )
not
1/((1/2) cos^2)
as 1/(1/2) is obviously 2.
Original post by mqb2766
1/2 sec^2 = 1/2 * 1/cos^2 = 1/(2cos^2 )
not
1/((1/2) cos^2)
as 1/(1/2) is obviously 2.

okay so i caved and looked at the answers
how the hell was i supposed to know not the break up tan but to break up sec 😓😓😓
https://imgur.com/a/9Mepb4H
Original post by esha06
okay so i caved and looked at the answers
how the hell was i supposed to know not the break up tan but to break up sec 😓😓😓
https://imgur.com/a/9Mepb4H

What you did was fine.
1/2 sec^2(x/2) / tan(x/2) = 1/2 cos(x/2)/(cos^2(x/2)sin(x/2)) = 1/(2cos(x/2)sin(x/2)) = cosec(x)
(edited 2 months ago)
Original post by esha06
okay so i caved and looked at the answers
how the hell was i supposed to know not the break up tan but to break up sec 😓😓😓
https://imgur.com/a/9Mepb4H

Your method was fine - you mixed up where your fractions was.in line 4 of the original.

If you have (1/2)something it means (something)/2 not (something)/ (1/2)