Question

differentiate please :D

Answer 1

y = -2/3cosx

Answer 2

Do you know the derivative of cosx?

Answer 3

I've tried it so many times and can't get the right answer

Answer 4

isn't it -sinx?

Answer 5

Correct. So then you can find the derivative of
\[\large y = \left( -\frac{ 2 }{ 3} \right) * (\cos x)\]keep in mind the -2/3 is just a constant,all you need to do is differentiate the cosx in the brackets.

Answer 6

so would the answer be 2sinx/3 ?

Answer 7

Correct.

Answer 8

Maybe the book is wrong because I have gotten so many different answers and I have no idea how the book got the answer;\[y=2cosx \div3\sin^2x \]

Answer 9

Was the problem \[\large y = -\frac{ 2 }{ 3 \cos x }\]

Answer 10

yep

Answer 11

Is it \(\frac{d}{dx}\left(-\dfrac2{3\cos x}\right)\)$?

Answer 12

Then... why not say something when i posted this, which is clearly not the same thing :P \[\large y = \left( -\frac{ 2 }{ 3} \right) * (\cos x)\]

Answer 13

well I just thought you had to do that...idk :P

Answer 14

You can't just take the cosx out of the denominator by magic... I wrote it that way because that's how y = -2/3cosx should be interpreted, without parentheses... make sure use them. It should be y = -2/(3cosx)

\[\large y = -\frac{ 2 }{ 3 \cos x } =\]
\[\Large y = \left( -\frac{ 2 }{ 3} \right) * (\cos x)^{-1}\]

You'll have to use the chain rule, but it's similar to what you did above.

Answer 15

I did \[-2\div3u ^{-1} \] where u = \[cosx\]

Answer 16

then it would equal \[2\div3u ^{-2} \times -sinx\]

Answer 17

then it would be 2/3(cosx)^-2 x -sinx

Answer 18

then i have no idea what to do :P or how it equals y=2cosx÷3sin^2x

Answer 19

\[\Large y = \left( -\frac{ 2 }{ 3} \right) * (\cos x)^{-1} \]
\[\Large \frac{ dy }{ dx } = - \left( -\frac{ 2 }{ 3} \right) * (\cos x)^{-2} ( - \sin x)\]
You can simplify a bit to:
\[\Large \frac{ dy }{ dx } = - \left( \frac{ 2 }{ 3} \right) * \frac{ \sin x }{ \cos x^{2} }\]

Answer 20

thanks heaps @agent0smith and sorry for the confusion :P

Answer 21

It doesn't match the answer in the book of \[y=2cosx \div3\sin^2x \] so i'd guess they made a mistake.

Answer 22

yeah they usually do... it's an old text book and I figured that's the right answer that you gave me :D

Answer 23

It is. You can confirm it with wolfram alpha... and they are not equivalent, either. http://www.wolframalpha.com/input/?i=-2sinx%2F%283cos%5E2x%29+%3D+2cosx%2F%283sin%5E2x%29+true+or+false