



Junior Member


Aug 13, 2009, 06:46 AM


miscellaneous problems in derivatives of trigonometric functions
please help me solve the problems below. As always, I end up with incorrect answers in some items of my reviewer. Here it is: find dy/dx
1.) y= xsinx + (11/2 x^2)cosx
Ans: 1/2 x^2 sinx
my ans: sinx + 1/2 x^2 sinx
2.) y = 1u/1+u ; u = cos2x
ans: 2tanxsec^2 x
my ans: (4sin2x)/(1+cos2x)^2
3.) x = cosu + usinu ;
y = sinu  ucosu
ans: tanu
my ans: (cosu + usinu)/(sinu + ucosu)
4.) y = tan(x+y)
ans: csc^2(x+y) = (1 + y^2)
my ans: sec^2(x+y)(1 + dy/dx)
hope you can help me! Thanks in advance. :)



Uber Member


Aug 13, 2009, 07:26 AM


I'll do the first one:
Break it separately;
Now, add both, you should be able to get the answer. :)
Just tell me if you're ready for the others.



Junior Member


Aug 13, 2009, 07:39 AM


Ah.. ok! Now I know what's wrong with my answer in number 1.. My derivative of xsinx is xcosx.. it should be xcosx + sinx..
I got it! Let's proceed to the next number.. :)



Junior Member


Aug 13, 2009, 07:49 AM


I already got the correct answer in number 2! :D



Junior Member


Aug 13, 2009, 07:50 AM


Oh no.. I mean I got the correct answer in number 3 and not in number 2. :)



Uber Member


Aug 13, 2009, 08:04 AM


Ok, y = 1u/1+u ; u = cos2x
That's good. Now, the proof. Expand the denominator;
You see that the answer include angles in x, not 2x? Well convert all the double angles to single ones.
To summarise;
See it? :)
Sigh, my mouse stopped working... :(



Junior Member


Aug 13, 2009, 08:24 AM


Ok! I got it.. I need to expand that to get the correct answer..
After few minutes, your mouse will start to work again. Be positive. :D
OK sir.. next number 4.. :)



Uber Member


Aug 13, 2009, 08:31 AM


Ok, for the last one then you did it right, just need to simplify.
Group the dy/dx:
Factorise dy/dx;
The cos^2(x+y) goes out;
Which is:
I think that's OK to end here. Furthermore, I've never done trigo differentiation that far... I mean, using implicit diff with trigo. Thanks for posting those questions. However, I don't know how to get the (1 + y^2).
And by the way, I'm 17, so I felt like embarrassed to be called 'sir', lol :o



Junior Member


Aug 13, 2009, 08:45 AM


haha.. okey! Thanks for the help! :D
I know there's a person here who knows that (1+y^2). :)



Junior Member


Aug 13, 2009, 08:47 AM


haha.. okey! Thanks for the help!
Forget about that 'sir' thing.. :D
I know there's a person here who knows that (1+y^2). :)



Uber Member


Aug 13, 2009, 08:47 AM


Oh, I just figured it out! Oh my God!!
Yay! LOL! :p



Junior Member


Aug 13, 2009, 09:06 AM


great! :D
hmmm.. I just want to know how
1/tan^2(x+y)  1 become 1/y^2  1... :)



Uber Member


Aug 13, 2009, 09:07 AM


Did you see earlier what y stands for?
Originally Posted by thinay
y = tan(x+y)



Junior Member


Aug 13, 2009, 09:12 AM


Oh.. I see.. tan^2(x+y) is same as (tan(x+y))^2.. So, when you break it down that will become (tan(x+y))(tan(x+y)). Since y=tan(x+y)... that will be (y)(y)=y^2..
I get it.. :D



Junior Member


Aug 13, 2009, 09:20 AM


wait... wait... last one.. I hope you can help me in this one.. :D
in here, not just finding the dy/dx.. but also the d^2y/dx^2..
here it goes..
y= sqrt of ((1cos4x)/(1+cos4x))
the answer in the 2nd derivative will be: 8sec^2 2x tan2x
This one, its really giving me a headache.. :(



Uber Member


Aug 13, 2009, 09:36 AM


OMG!! I don't know where the modulus came...
Phew!!
Oh my... did you get that?



Uber Member


Aug 13, 2009, 09:41 AM


Now;
Now, can you find the second derivative?



Junior Member


Aug 13, 2009, 09:43 AM


Actually not.. its solution is very long.. and I get confused because of that.. I will go back to the very start and review my solution.. In the end, I end up with nothing.. Still confused on how to get its derivative.. :))



Junior Member


Aug 13, 2009, 09:45 AM


Oh.. I see.. I'll be the one to get the second derivative.. thanks for the help.. Again and again.. :D



Uber Member


Aug 13, 2009, 09:47 AM


A simple 'rate this answer' to my post would be the least you can do! :) I'm trying to collect 'good ratings' ;) :p


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