Incorrect answers to derivative of this equation?

raven2k7 said:
is that some kind of shortcut method ? cause the last part confused me where u put (1/2 *) what was in the bracket. did you cancel the 2s out?
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Please go through the solution shown in response (3). If you are still stuck show us how far did you go and point out where did you get stuck.
 
BraveHeart258 said:
Here is the answer:
these are so much fun! :)
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Ln( A-B)[MATH]\neq[/MATH]Ln(A) - Ln(B) so your working is not correct. Besides your final answer of -csc does not even make any sense.
Also sinx/cosx [MATH]\neq[/MATH]cscx, it equals tanx
More importantly why would you think that giving a complete answer would be a positive thing to do on a math help forum?
 
Since it was recommended that the OP goes through post #3, which is not valid at all I will show the 1st few steps.

[MATH]f(x) = \ln \sqrt {\dfrac{1+cosx}{1-cosx}} = \dfrac{1}{2}\ln(\dfrac{1+cosx}{1-cosx})=\dfrac{1}{2}[\ln(1+cosx)-\ln(1-cosx)][/MATH]
Now take the derivative of both sides,
 
Jomo said:
Since it was recommended that the OP goes through post #3, which is not valid at all I will show the 1st few steps.

[MATH]f(x) = \ln \sqrt {\dfrac{1+cosx}{1-cosx}} = \dfrac{1}{2}\ln(\dfrac{1+cosx}{1-cosx})=\dfrac{1}{2}[\ln(1+cosx)-\ln(1-cosx)][/MATH]
Now take the derivative of both sides,
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The recommendation was made with a hope that the original poster will ask those questions (Is this step correct? Why?). It was NOT recommended to be taken as correct.
 
LN() and Log() act the same with regards to using the exponential

I sorta figured it out when the teacher wrote it on the board or asked him questions. I had to look up my old equations. I spent like 1 hour thinking and researching... it must have been 5 years and 7 years since I last did this. Best part of my Calculus book was the equations on the last three pages on the back. They are lifesavers.

Ln(x2)= 2Ln(x) is similar to Log(x2)=2Log(x)

I love giving out the answer! It makes them so happy! Plus I am a reverse thinker. Sometimes I need to see the problem worked out to know how to do it myself. But its fun the other way around and its better for your grow. You need think about what you are doing before writing. Sometimes, I all needed was to read the beginning of the chapter and really try to understand it and take it in.
 
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Subhotosh Khan said:
The recommendation was made with a hope that the original poster will ask those questions (Is this step correct? Why?). It was NOT recommended to be taken as correct.
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I knew that was what you meant. I however felt that the OP would have taken the post as being correct. Based on their last post I was correct.

I always put the student first even if it makes me look bad to a helper. Please understand that my motto has been and will always be "The students come first"
 
Jomo said:
I knew that was what you meant. I however felt that the OP would have taken the post as being correct. Based on their last post I was correct.

I always put the student first even if it makes me look bad to a helper. Please understand that my motto has been and will always be "The students come first"
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I never doubted that.
 
firemath said:
Oh, boy. Why do I feel like this forum is not your style.....?
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Jomo said:
I knew that was what you meant. I however felt that the OP would have taken the post as being correct. Based on their last post I was correct.

I always put the student first even if it makes me look bad to a helper. Please understand that my motto has been and will always be "The students come first"
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okay I am sorry. It won't happen again. I will ask them questions on how to make the students think about the direction of the problem
 
Jomo said:
Since it was recommended that the OP goes through post #3, which is not valid at all I will show the 1st few steps.

[MATH]f(x) = \ln \sqrt {\dfrac{1+cosx}{1-cosx}} = \dfrac{1}{2}\ln(\dfrac{1+cosx}{1-cosx})=\dfrac{1}{2}[\ln(1+cosx)-\ln(1-cosx)][/MATH]
Now take the derivative of both sides,
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Can you explain to me why my direction was invalid?
 

Hi Raven, I see a few mistakes in your work which I also made mistakes of myself. I noticed that cosecant wasn't properly described? make sure to use -csc(x)dx defining that you took the derivative.

A teacher once explained to me in AP Calculous 1 (dx) was an infinitesimal small rectangle that allows us understand there is a change from derivative of the function in respect to the unknown variable in this case x. Make sure you write the following after [MATH]-csc(x)\dfrac{df}{dx}[/MATH] . Some mathematicians will remove [MATH]df[/MATH] all together and just ask you to put [MATH]dx[/MATH] or [MATH]\dfrac{d}{dx}[/MATH].
Most of the time it is more proper if make this notation to the function changed of the [f(x)] or a variable such a y into the change of the of the variable unknown in this case x. You will most like see problems with the following notation [MATH]\dfrac{dy}{dx}[/MATH]I am not sure maybe its a matter of controversy in the math world.

If you take a derivative on one side of the equation, you take the derivative on the other side so [MATH]-\dfrac{1}{sin(x)}\dfrac{df}{dx}[/MATH]
Also make sure to call out csc(x) since the function is based on the unknown variable of x. The book must have made a mistake since there is not [MATH]\dfrac{df}{dx}[/MATH] after the csc(x) making it not [MATH]-csc(x)\dfrac{df}{dx}[/MATH]
I am looking at your answer in post #18 and I do not know how you removed function of ln() from the equation. You have to take the derivative of ln() in order to remove ln() from the equation. You must show more steps between line 1 or line 2. If you jump ahead, in your work you might end up problems with making other problems. its better to layout clearly and make sure you are not jumping steps.

These issues are related to Physics and making sure you have proper units. Learn this now to avoid having issues in other classes.

Don't feel hard on yourself because I made the same errors in my answer having no x and no dx. We all make mistakes. Also the site I was using should have used the following equation instead [MATH]-\dfrac{1}{f(x)}*f'(x)\dfrac{df}{dx}[/MATH]
 
What are you talking about??

The final answer should be [MATH]-\csc(x)[/MATH], not just [MATH]-\csc[/MATH]; but there's nothing else missing there. The goal is to find [MATH]f'(x[/MATH]), which is the same thing as [MATH]\frac{df}{dx}[/MATH].
 
notation was missing.

[MATH]\frac{df}{dx}[/MATH] and [MATH]-\csc(x)[/MATH] and the student jumps a lot of steps between line 1 and line 2. It is clearly correct but doesn't describe when he is taking the derivative of the function in post #18

Jumo called out my answer as incorrect when it clearly is correct with minor flaws similar to how the student performed. I don't want to mislead the student.
 
That is incorrect I provided the formula for what I used in the equation.

Jomo said:
Sure. I actually already did. You used that ln( A - B) = ln (A) - ln (B) which is not correct.
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I clearly have it written have the equation on the right side of the paper. I ways always told by college math teachers, physics, chemistry to state the equation or law before performing it.

Equation used:

[MATH]ln(\dfrac{A}{B})=ln(A)-ln(B)[/MATH]
 
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