Question

Derive log(sin(x)) by rearranging.

I get up to dy/dx = ln(10)sin(x)/cos(x)

Answer 1

Could you explain how you found dy/dx = ln(10)sin(x)/cos(x). What steps did you take?

Answer 2

I get up to
\[\frac{dy}{dx}=\frac{\ln(10)\sin(x)}{\cos(x)}\]

I don't know why I'm getting the inverse of the answer. I will type the steps

Answer 3

OK, you are very close to the right answer! :) Let's take a look at your steps...

Answer 4

y=log(sin(x))

\[10^y=sinx\]
\[\frac{d *ln(10)*10^y}{dy}=\frac{d* cosx}{dx}\]
\[\frac{dx}{dy}=\frac{cosx}{ln(10)*10^y}\]
then more...

Answer 5

Hold on one moment

Answer 6

\[\frac{dy}{dx} =\frac{\ln(10)*\sin(x)}{\cos(x)}\]

Answer 7

I'm making some stupid mistake but I don't know where

Answer 8

I understood everything up to 10^y = sin(x) --> d*ln(10)*10^y/dy = d * cos x / dx

Answer 9

Is that step valid?

Answer 10

Oh, wow, I did a stupid :P. I've never taken the derivative of both sides to solve before. Is there a valid step that you can take to end up with dy/dx (without using the chain rule)?

Answer 11

oh

Answer 12

Wait

Answer 13

It should be:

Answer 14

\[\ln(10)10^y*dy=\cos(x)*dx\]

Answer 15

Is that right?

Answer 16

Well, yes, but you are still using the chain rule to solve it. But it is mathematically correct. :) There is another way if you need it.

Answer 17

You've used a sort of "sneaky" chain rule :)

Answer 18

Haha, yeah, I can kind of see that :)

What is this other way you speak of? :P

Answer 19

Do you know the log conversion:
\[\log_a(b) = \frac{\ln{b}}{\ln{a}}\]

Answer 20

Have you seen/used that before?

Answer 21

Yeah, just a sec, I'll try and work out how to use it myself and if not I will ask you what the method is. :)

I like trying to work things out first.

Answer 22

A very good habit. I applaud you! :D

Answer 23

Does it also use the "sneaky" chain rule?
I can only see a way to do it with the same steps but with 'e' as the base using that conversion, which makes the 'dy' slightly easier. Is that what you were thinking of?

Answer 24

Ah, you're right! Let me see if I can cook up any other methods...

Answer 25

Haha, okay cool. I like it! And that's all right, I've got three now. I really like the idea or rearranging and taking the derivative of both sides compared to the set chain rule, for the simpler ones at least.

Thanks for your help :)

Answer 26

No worries, you did very well... I'll have to think about this one for future problems. :)

Answer 27

Haha, thanks