1. ## Derivate.

given is $\displaystyle y(x)=ln(tan(\frac{x}{2}))-atan(x)ln(1+sin(x))-x$

so I would like to derivate this

thus we can split it in 3 part

First $\displaystyle ln(tan(\frac{x}{2}))$ wil be $\displaystyle \frac{1}{2}\frac{sec^2(\frac{x}{2})}{tan(\frac{x}{ 2})}$

the second and third one $\displaystyle atan(x)ln(1+sin(x))-x$ will be $\displaystyle \frac{ln(1+sin(x))}{sin^2(x)}-\frac{atan(x)cos(x)}{1+sin(x)}-1$

finelay we get $\displaystyle \frac{1}{2}\frac{sec^2(\frac{x}{2})}{tan(\frac{x}{ 2})} - \frac{ln(1+sin(x))}{sin^2(x)}-\frac{atan(x)cos(x)}{1+sin(x)}-1$

Who can I simplify this? Greets Thanks a lot.

2. Hello, Bert!

An error in your derivative (probably a typo)
. . Otherwise, excellent work!

Differentiate: .$\displaystyle y(x)\;=\;\ln(\tan\frac{x}{2}) - \arctan(x)\cdot\ln(1 + \sin x ) - x$

You should get: .$\displaystyle \frac{1}{2}\cdot\frac{\sec^2\frac{x}{2}}{\tan\frac {x}{2}} - \frac{\ln(1+\sin x)}{\underbrace{1 + x^2}}-\frac{\arctan(x)\cdot\cos x}{1 + \sin x }-1$

With three (four!) different denominators, there's not much chance of simplifying.

The first term can be simplified a bit:

. . $\displaystyle \frac{1}{2}\cdot\frac{\frac{1}{\cos^2\frac{x}{2}}} {\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}}\;=\; \frac{1}{2\sin\frac{x}{2}\cos\frac{x}{2}}$ $\displaystyle = \;\frac{1}{\sin x} \;=\;\csc x$ . . . but that's all . . .

3. oké but the answer in my book say $\displaystyle f'(x)=\frac{ln(1+sin(x))}{sin^2(x)}$

What is the problem ? Greets Thanks .

4. Hello again, Bert!

but the answer in my book say $\displaystyle f'(x)=\frac{\ln(1+\sin x )}{\sin^2 x }$

What is the problem ?

Simple . . . the book is wrong!

I'm puzzled . . . you got the same wrong answer.
. . Did you just copy the book's answer and trust it to be correct?

Are you aware that the derivative of $\displaystyle \arctan x$ is: .$\displaystyle \frac{1}{1+x^2}$

There's no way to get $\displaystyle \sin^2x$ in the denominator
. . and you should know that.

5. Hello

of course not I don't trust my book but the problem is if I integrate it by derive on my computer he tells me that integrate of $\displaystyle \frac{ln(1+sin(x))}{sin^2(x)}$

will be $\displaystyle ln(tan \frac{x}{2})-cot(x)ln(sin(x)+1)-x$

here a screen shot of my window

Greets.

6. Hello, Bert!

The problem is if I integrate it by derive on my computer

it tells me that: .$\displaystyle \int\frac{\ln(1+ \sin x )}{\sin^2x}\,dx \;= \;\ln\left(\tan\frac{x}{2}\right) - \underbrace{\cot x}\cdot\ln(\sin x +1) -x$

I don't see a problem . . .
That is not the original function: .$\displaystyle \arctan x\cdot\ln(1 + \sin x)$

7. maybe the problem is that we define in an other way cotan I mean $\displaystyle \frac{1}{tan}$

you use arctan I think thats the function who give back an angle I don't use this.

Maybe I copy the exercise at a wrong way so let me show an excat copy.

Greets Tanks a lot.

8. Hello, Bert!

Sheesh! . . . That explains our different answers . . .

The function is: .$\displaystyle y \;= \;\ln\left(\tan \frac{x}{2}\right) - \underbrace{\cot x}\cdot\ln(1 + \sin x) - x$

We know that: .$\displaystyle \tan^3x$ means $\displaystyle (\tan x)^3$

. . but $\displaystyle \cot x \:=\:\frac{1}{\tan x}$ and should be written $\displaystyle (\tan x)^{-1}$ **

The symbols .$\displaystyle \arctan x,\;\tan^{-1}x\text{, and your {\bf atan} }x$
. . are used to signify the inverse tangent of $\displaystyle x$

That is, $\displaystyle \text{asin }x$ does not mean $\displaystyle \frac{1}{\sin x}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**
I don't understand why you replaced $\displaystyle \cot x$ with its reciprocal.
. . There is a formula: .$\displaystyle (\cot x)' \;= \; -\csc^2x$

9. this will be the problem I think The matter is that ther are different notations in different language area.

So let we work with the original copy of my book.

Where do I miss then in my answer?

Greets Thanks.

10. Hello again, Bert!

I got it! . . .
They did more simplifying than I thought humanly possible!

We have: .$\displaystyle y \;= \;\ln\left(\tan\frac{x}{2}\right) - \cot x\cdot\ln(1 + \sin x) - x$

Then: .$\displaystyle y' \;= \;\frac{1}{2}\cdot\frac{\sec^2\frac{x}{2}}{\tan \frac{x}{2}} -$$\displaystyle \left[\cot x\cdot\frac{\cos x}{1 + \sin x} - \csc^2x\cdot\ln(1 + \sin x)\right] - 1$

I've already shown that the first fraction can be simplfied.

. . So we have: .$\displaystyle y' \;= \;\csc x - \frac{\cot x\cdot\cos x}{1 + \sin x} + \csc^2x\cdot\ln(1 + \sin x) - 1$

The second fraction is: .$\displaystyle \frac{\frac{\cos x}{\sin x}\cdot\cos x}{1 + \sin x} \;=$ $\displaystyle \;\frac{\cos^2x}{\sin x(1 + \sin x)}\;= \;\frac{1 - \sin^2x}{\sin x(1 + \sin x)}$

. . . $\displaystyle =\;\frac{(1 - \sin x)(1 + \sin x)}{\sin x(1 + \sin x)} \;= \;\frac{1 - \sin x}{\sin x}\;=$ $\displaystyle \frac{1}{\sin x} - \frac{\sin x}{\sin x}\;=\;\csc x - 1$

The third term is: .$\displaystyle \csc^2x\cdot\ln(1 + \sin x)\;=\;\frac{\ln(1 + \sin x)}{\sin^2x}$

. . So we have: .$\displaystyle y' \;= \;\csc x -(\csc x - 1) + \frac{\ln(1 + \sin x)}{\sin^2x} - 1$

Therefore: .$\displaystyle \boxed{y' \;= \;\frac{\ln(1 + \sin x)}{\sin^2x}}$ . . . ta-DAA!

11. thank you very much I see.

Greets.