Find the derivative:

- anonymous

Find the derivative:

- katieb

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

|dw:1349927470125:dw|

- anonymous

I can easily find the derivative of the ln(x+ln(x+ln(x))) but I cant figure out how to find the derivative of the first part.

- anonymous

i think it will be better to take the log of that part with the respect of e base.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

|dw:1349927833215:dw|

- anonymous

|dw:1349927846691:dw|

- anonymous

yeah.

- anonymous

Okay I had an idea about that. But don't you have to do that for the ln(x+ln(x+ln(x))) as well?

- anonymous

Like I thought you had to take the log of both sides for logarithmic differentiation.

- anonymous

yes,or you may solve the two parts one by one

- anonymous

So I thought you had to:|dw:1349927977970:dw|

- anonymous

Do you do that?

- anonymous

Don't*

- anonymous

and even if you take the ln of ln(x+ln(x+ln(x))),there ain't any problem

- anonymous

Yes yes. I knew that. But is that valid? I though you had to Log both sides of the ENTIRE term.

- anonymous

Ohh? Why not?

- anonymous

yes it is.

- anonymous

How is it valid? Sorry for being stubborn.

- hartnn

sorry to interrupt. but ln(A+B)\(\ne\)ln A + ln B
so here you have to differentiate 1st term separately , using ln

- anonymous

Ohh I know @hartnn . I didn't seperate it.

- anonymous

|dw:1349928327564:dw|

- anonymous

that is why you can find the derivatives separately

- anonymous

Lol I know. By I mean why can you ln only the (arcsin)^Tan(x)? I though you had to ln EVERYTHING.

- hartnn

"had to" <---no

- anonymous

why?? if you solve them part by part, you are free to take the ln of one part without changing the other

- anonymous

Ohh. Okay. That explains it.

- anonymous

Fair enough haha...

- anonymous

Could I do this instead?
|dw:1349928628192:dw|

- anonymous

For the derivative I mean.

- anonymous

@sritama ,@hartnn

- anonymous

@hartnn

- hartnn

NO!
what have u done!!

- anonymous

i couldn't get it @Dido525 :(

- anonymous

You make me chukle @hartnn .

- anonymous

how and why will you do that?

- anonymous

Well the derivative of a^x = a^x *ln(a)

- hartnn

ln y = tan x ln(sin^-1 (x))
diff. this.
no other way

- anonymous

Okay okay :P .

- anonymous

Then this question isn't so hard.

- hartnn

is it ?
what u get as y' ?

- anonymous

Wait.

- anonymous

Doing it now.

- anonymous

I got:
|dw:1349929346410:dw|

- anonymous

For the derivative of (sin-1(x))^(tan(x))

- anonymous

@hartnn

- hartnn

seems correct.

- anonymous

Okay :D . Thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.