Find dr/dt for r=(csc(t)+cot(t))^-1

I tried chain rule and got

-(csct+cott)^-2*-csct+cott+(-csc^2t)

Anyone see my error?

Question

Find dr/dt for r=(csc(t)+cot(t))^-1

I tried chain rule and got

-(csct+cott)^-2*-csct+cott+(-csc^2t)

Anyone see my error?

anonymous · Answer

no

kmeis002 · Answer

check the derivative of $csc(t)$, plus be careful with your parenthesis.

welshfella · Answer

i cant remember the derivatives of cosec and cot...

kmeis002 · Answer

and your addition signs, I'm not sure if you meant to put it there or if there's a typo.

idku · Answer

I would re-write in sines and cosines in the parenthesis combine and then differentiate using quotient,.

anonymous · Answer

deriv of csc(t) is -csc(t)*cot*(t) I don't think that part is incorrect. Are my parenthesis effecting my answer?

welshfella · Answer

derivative of cscc x = -cot x csc x   and of cot x is - csc^2 x

kmeis002 · Answer

You have it is $-csc(t)+cot(t)$

welshfella · Answer

* csc x

welshfella · Answer

yes the whole of the second part should be in parentheses

anonymous · Answer

Thank you so much! I suck at this stuff

welshfella · Answer

yw 
- it needs some practice to get confident in the more complicated derivatives

anonymous · Answer

Some calculations using Mathematica 9 are attached.