Question

help pls....

Answer 1

if
\[y = cosx \times cosec^{n-1}x\]
then prove
\[\frac{ dy }{ dx } = (n -2) cosec^{n - 2}x - ( n -1)cosec^{n}x\]

Answer 2

@phi , @hartnn , @ganeshie8 , help!!

Answer 3

what does the n stand for?; otherwise this just seems like derivatives

Answer 4

n is a constant and yes it's all about derivatives and differentiation but i can't remove the "cosx" and "sinx" which i get when i differentiate this one

Answer 5

the answer only have "cosec x" in it but i get "sin x " , "cos x" , "cot x" in my answer .... wt to do ?

Answer 6

\[cos * csc^n\]
its a product rule ...

\[cos' * csc^n+cos * csc'^n\]
\[-sin * csc^n+cos * n~csc^{n-1}~-csc~cot\]
right?

Answer 7

ok ! . so.....

Answer 8

now we can turn it all into sin cos parts
\[-\frac{sin}{sin^n}-cos * n~sin^{1-n}~\frac{1}{sin}~\frac{cos}{sin}\]
\[-\frac{1}{sin^{n-1}}-n~cos^2 ~sin^{1-n-2}\]
\[-\frac{1}{sin^{n-1}}-n~(1-sin^2) ~sin^{-n-3}\]

Answer 9

yeah... so far is good to me :)

Answer 10

im going to replace n with n-1 just to make it more inline with the actual problem; but now that we have all sins, its looking promising to me.

\[-\frac{1}{sin^{n-1-1}}-(n-1)~(1-sin^2) ~sin^{-(n-1)-3}\]
\[-csc^{-n}-(n-1)~(1-sin^2) ~sin^{-n-2}\]
\[-csc^{-n}-(n-1)~(sin^{-n-2}-sin^{-n-2+2}) \]
\[-csc^{-n}-(n-1)~(csc^{n+2}-csc^{n}) \]
\[-csc^{-n}-(n-1)csc^{n+2}+(n-1)csc^{n} \]

im sure i got some typo in this someplace....

Answer 11

hmmmm..... is this all ?

Answer 12

but I think I got the trick here ... thanks you a lot !! :)

Answer 13

thats the process, but without the typos accumulating along the way :)