Question

Prove that d/dx (csc x)=-csc x cot x

Answer 1

that's already as the general formula

Answer 2

well,
y = csc x = 1/sinx

use the quotient rule :
y ' = (vu' - uv')/v^2

Answer 3

right, but I am needing to prove it with quotient rule... so basically I get : (sinx)(1)'-(1)(sinx)' and all of that over (sinx)^2

Answer 4

thank you for response, okay so i have gotten that far..

Answer 5

so can i do anything after I fill in for quotient rule?

Answer 6

yep, you are right
y = 1/sinx
u = 1 ---> u' = 0
v = sinx ---> v' = cosx

y' = (vu' - uv')/v^2 = (sinx * 0 - 1 * cosx)/(sinx)^2 = (0 - cosx)/sin^2 x
= -cosx/(sinx*sinx) = -cosx/sinx * 1/sinx = -cotx cscx

Answer 7

note :
-csc x cot x = -cot x-csc x :)

Answer 8

oppss. typo
note : -csc x cot x = -cot xcsc x :)

Answer 9

THANK YOU!!! :))

Answer 10

I HAVE A FEW MORE like this if you wanna help!
i started them but only got half way through i think

Answer 11

d/dx [e^x (tanx-x)]
Differentiatie each function.

Answer 12

i ended up getting (e^x)(sec2x-1) + (tanx-x)(e^(x-1))

Answer 13

derivative of e^x just still e^x not e^(x-1)