Question

Hi there, trig derivative question. I am asked to find the second derivative of y'' if y=cscx and if y=secx. I run it out and come to (cscx)^3+cscx(cotx)^2 (for y=cscx) and I think I can finagle it down to y''=-cscx, but the back of the book is telling me that y''=2(cscx)^3 -cscx. Does anybody know what I'm missing? I've checked my signs and all, am I just missing an identity? Am I trying too hard, and its actually super easy? Thanks in advance.

Answer 1

what did you get for d/dx (csc(x))?

Answer 2

I got cosecx + 2cos^2(x)/sin^3(x)

Answer 3

Hey Christopher :)
Were you converting to sines and cosines and all that business to take your derivative?
Or you have your shortcuts for cscx derivative handy?

Answer 4

\[y=\csc x\]
\[y'=-(\csc x)(\cot x)\]
Then
\[y''=\csc ^{3}x+\csc x \cot ^{2}x\]
Is how I got to an unsimplified answer (using product rule) which can be reduced as @alekos said as well.

Answer 5

Ok looks great so far! :)\[\Large\rm y''=\csc ^{3}x+\csc x \color{orangered}{\cot ^{2}x}\]\[\Large\rm y''=\csc ^{3}x+\csc x \color{orangered}{(\csc^2x-1)}\]

Answer 6

I was just using the rules for trig differentiation, zepdrix

Answer 7

You do have the right answer. Im confused, what are you asking exactly?

Answer 8

Oh I see, did you accidentally put 1-csc^2?

Answer 9

Ahh, I was using the identity wrong! Thanks for the nudge in the right direction, I had \[\csc ^{2} x + 1\] mistakenly. I'll try the correct one now!

Answer 10

cool :)

Answer 11

Out of curiosity, do you know if it is possible to get down to y''=-csc x, or was that just the mad attempts of an over-mathed mind?

Answer 12

Mmm I think the crazy night time math is gettin to yer brain.
I don't see any way we could widdle that 2csc^3x down to nothing.

Answer 13

Upon further review, I based that assumption on an eraser mark stuck to the paper in the shape of a negative sign. That'll teach me to math without glasses!

Answer 14

lol XD