Question

I quite honestly am not sure how to start this one...

Find y' if 3 = (cos(x))^sin(y).

Any and all help is greatly appreciated!

Answer 1

take ln( ) of both sides

Answer 2

this will allow you to bring that sin(y) down by using the power rule for log

Answer 3

Is this an Implicit Differentiation situation?

Answer 4

it is an implicit equation so yes

Answer 5

Thank you! Alright... I'm gonna dive in, and see what I can get as an answer!

Answer 6

Okay, so I got:
(and a snack)

\[\frac{1}{3} = \sin(y)*(-\tan(x))\]

Answer 7

So... what's "y prime" if it doesn't start off equal to y?

Answer 8

No, I don't think that's quite right

Answer 9

@amonoconnor do you know chain rule?

Answer 10

also ln(3) doesn't equal 1/3
and it looks like you completely missed that there was a product on the right hand side

Answer 11

oh but it being ln(3) or 1/3 won't matter in the end
because the derivative of ln(3) or even 1/3 is 0
but yeah it should be ln(3) on the left hand side before differentiating
and then 0 after differentiating

Answer 12

Is this right?
\[3 = (\cos(x))^{\sin(y)}\]
\[\ln3 = \ln[\cos(x)^{\sin(y)}]\]
\[\ln3 = \sin(y) * \ln(\cos(x))\]
Now taking the derivative...
\[(d/dx(\ln3) = 1/x = 1/3) = (\cos(y)) + (\frac{1}{\cos(x)})*(-\sin(x))\]
\[\frac{1}{3} = \cos(y) + (-\tan(x))\]
\[1/3 = \cos(y) - \tan(x)\]

Answer 13

but d/dx(ln(3)) is 0
you cannot treat 3 like it is x ...

Answer 14

so... \[d/dx(lnx) = \frac{1}{x}\]

does't apply?

Answer 15

also your chain rule part is missing when you differentiated sin(y)
you also did not bring down the parts of the product rule
you are for some reason saying (fg)'=f'+g'
while this is not true
(fg)'=f'g+g'f

Answer 16

3 is not x
3 is a constant

Answer 17

d/dx (constant)=0
d/dx( ln(3))=0 since ln(3) is constant

Answer 18

Gotcha... okay, let me try again. :)

Answer 19

\[\ln(3)=\sin(y) \cdot \ln(\cos(x)) \\ \frac{d}{dx}(\ln(3))=\frac{d}{dx}[ \sin(y) \cdot \ln(\cos(x)] \]
try using constant rule on left
and product rule on right hand side
and for both factors in the product you will need to use chain rule
you are differentiating ln(cos(x)) correctly
but not sin(y)

Answer 20

\[\frac{d}{dx} \sin(y)=y' \cos(y)\]

Answer 21

\[y' = \frac{\sin(y)\tan(x)}{\cos(y)*\ln(\cos(x))}\]

Answer 22

\[0=\ln(\cos(x) ) \cdot \frac{d}{dx} (\sin(y))+ \sin(y) \cdot \frac{d}{dx}(\ln(\cos(x)) \\ 0 =\ln(\cos(x)) \cdot y' \cos(y)+\sin(y) \cdot \frac{-\sin(x)}{\cos(x)}\]
and then solve for y'
this will involving adding something on both sides
then dividing something on both sides


oh wow nevermind
looks likes you did it

Answer 23

looks awesome

Answer 24

Thank You!!!

Answer 25

you could do one more thing

Answer 26

but you are answer is right

Answer 27

What else!? :)

Answer 28

\[y'=\frac{\sin(y) \tan(x)}{\cos(y) \cdot \ln(\cos(x))}= \frac{\frac{\sin(y)}{\cos(y)} \tan(x)}{\ln(\cos(x))} =\frac{\tan(y) \tan(x)}{\ln(\cos(x))}\]

Answer 29

but you don't have to do that

Answer 30

do you have any questions on this question/

Answer 31

Nope, not anymore:) Thank you again!

Answer 32

kk cool stuff

Answer 33

well done connor

Answer 34

Thank you! :)