Question

what is the derivative of sec^2y + cot^2x = 3?

Answer 1

I got this right but i just got half-credit because the choices utilized identities

Answer 2

Well let's see... show me what you did.

Answer 3

sec^2(y)tan^2(y) -csc^2(x) = 0

Answer 4

but from there on out. my prof labeled me wrong T.T

Answer 5

the answer is cotxcsc^2x / sec^2ytany

Answer 6

Well naturally :)
Derivative of sec²(y) is 2sec²(y)tan(y) y'

Answer 7

Let's differentiate \(\large \sec^2(y)\) using chain rules:
\[\Large \implies \color{red}{2\sec(y)} \frac{d}{dx}\sec(y)\]\[\Large \implies 2\sec(y)\color{red}{\sec(y)\tan(y)}\frac{d}{dx}y\]\[\Large \implies 2\color{red}{y'}\sec^2(y)\tan(y)\]

gets? ^_^

Answer 8

oh look at that cheese! haha!

Answer 9

where I really went wrong is with the identities

Answer 10

You'd go wrong with them if you went wrong with differentiation in the first place.
Now, did you understand the proper way to differentiate sec²(y) ?

Answer 11

yeah. I messed up

Answer 12

So we now have
\[\Large 2\color{red}{y'}\sec^2(y)\tan(y )-2\cot(x)\csc^2(x)=0\]

Answer 13

Do I have to go through the nitty gritty of differentiating cot²(x) ?

Answer 14

no need
i recognize my mistake

Answer 15

wrong grammar pa haha

Answer 16

hmm...

Answer 17

2y′sec2(y)tan(y))=2cot(x)csc2(x)

Answer 18

2y′=2cot(x)csc2(x)/sec2(y)tan(y))

Answer 19

y'=cot(x)csc2(x)/sec2(y)tan(y)

Answer 20

mhmm :)
\[\Large \color{red}{y'}= \frac{\cot(x) \csc^2(x)}{\tan(y)\sec^2(y)}\]

Answer 21

i fluttering hate myself for not getting this correct >.< rookie mistake

Answer 22

Ahh don't worry about it ^_^

Answer 23

We all make mistakes XD