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Madasmaths very hard question

Prove that the differential of 1+coshx/1-coshx is equivalent to coth(x/2)Cosech^2(x/2)

Would very much appreciate if anyone could offer a solution or a way to start from the RHS for this question. Its question 4 of part 4 of his differentiation practice iii booklet.
Original post by DatGuyWelbz23
Prove that the differential of 1+coshx/1-coshx is equivalent to coth(x/2)Cosech^2(x/2)

Would very much appreciate if anyone could offer a solution or a way to start from the RHS for this question. Its question 4 of part 4 of his differentiation practice iii booklet.


All you need to do is differentiate 1+coshx1coshx\dfrac{1+\cosh x}{1-\cosh x} and show that this result is the same as cothx2cosech2x2\coth \dfrac{x}{2} \mathrm{cosech}^2 \dfrac{x}{2}.

To differentate, you should employ the quotient rule. What do you get?
Original post by RDKGames
All you need to do is differentiate 1+coshx1coshx\dfrac{1+\cosh x}{1-\cosh x} and show that this result is the same as cothx2cosech2x2\coth \dfrac{x}{2} \mathrm{cosech}^2 \dfrac{x}{2}.

To differentate, you should employ the quotient rule. What do you get?

-2sinhxcoshx/(1-coshx)^2. Unsure of how to progress from that stage. From inspection i presumed multiplying through the LHS by 1+coshx would be whats required, but couldnt progress that way either.
Original post by DatGuyWelbz23
-2sinhxcoshx/(1-coshx)^2. Unsure of how to progress from that stage. From inspection i presumed multiplying through the LHS by 1+coshx would be whats required, but couldnt progress that way either.


Might want to double check your numerator. It's not correct.
Alright, cheers for the help mate.
Anybody know how to start from the right hand side?
Original post by DatGuyWelbz23
Anybody know how to start from the right hand side?


It's more obvious to go from LHS to RHS. I.e. from 2sinhx(1coshx)2\dfrac{2\sinh x}{(1-\cosh x)^2} to cothx2cosech2x2\coth \dfrac{x}{2} \mathrm{cosech}^2 \dfrac{x}{2}.

Once you do that, then to go from RHS to LHS you just work backwards.

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