Question

find the derivative of: f(t)=9tan(t)+7csc(t)

my answer is: 9sin3x-7cos2x all over cosxsin3x

just wondering if thats right.

Answer 1

[ 9tan(t) ]' = 9/cos^2(t)
[7ccs(t)]' = -7cos(t)/sin^2(t)

Answer 2

9⋅(sec(t))^2−7⋅cot(t)⋅csc(t)

Answer 3

yes same as i wrote

Answer 4

so the only thing i have wrong is that i need to put cos2 in the denominator?

Answer 5

sin3x is sin^3(x) for you ?

Answer 6

yes

Answer 7

so it still doesnt make it right since the second term will be missing stuff

Answer 8

well for my derivative i have
9sec^2(x)+7(-csc2x)cot^2(x)

Answer 9

9/cos^2(t) - -7cos(t)/sin^2(t) =

[9sin^2(t) - 7cos^3(t)] / sin^2(t)cos^2(t)

Answer 10

see i had sin^3 in my denominator

Answer 11

so there is something wrong with the second term since :

[7ccs(t)]' = -7cos(t)/sin^2(t)
or we can represent it as
-7cot(t)csc(t)

Answer 12

yes that's what i have, then i changed it all into sin and cos.

Answer 13

but you wrote 7(-csc2x)cot^2(x)

Answer 14

this is not the same

Answer 15

i dont know why i wrot that hah.

Answer 16

lol so show me your answer before manipulation

Answer 17

now my answer is: \[\frac{ 9\sin ^{2}x-7\cos ^{2}x }{ \cos ^{2}xsin ^{2}x }\]

Answer 18

still wrong

Answer 19

is that correct?

Answer 20

you should have 9sin^2(x) -7cos^3(x) upstairs

Answer 21

but the bottom is correct? i see what i did wrong on the top now when i got my lcd i didnt add the other cos x to the 2cos^2(x)

Answer 22

the bottom correct but the numerator isnt

Answer 23

so my numerator should be 9sin^2(x) -7cos^3(x)

Answer 24

yes

Answer 25

and it should be t instead of x ;)

Answer 26

so done :) ?