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thinay · Aug 13, 2009, 06:46 AM

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 + (1-1/2 x^2)cosx
Ans: 1/2 x^2 sinx
my ans: -sinx + 1/2 x^2 sinx

2.) y = 1-u/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. :)

Unknown008 · Aug 13, 2009, 07:26 AM

I'll do the first one:

Break it separately;

$\frac{d(x\,sin(x))}{dx} = (x)(cos(x)) + (sin(x)) = xcos(x) + sin(x)$

$\frac{d((1-\frac{x^2}{2})cos(x))}{dx} = (1+\frac{x^2}{2})(-sin(x)) + (cos(x))(-\frac{2x}{2}) = -xcos(x) -sin(x) + \frac{x^2}{2}sin(x)$

Now, add both, you should be able to get the answer. :)

Just tell me if you're ready for the others.

thinay · 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.. :)

thinay · Aug 13, 2009, 07:49 AM

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

thinay · Aug 13, 2009, 07:50 AM

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

Unknown008 · Aug 13, 2009, 08:04 AM

Ok, y = 1-u/1+u ; u = cos2x

$\frac{dy}{dx} = \frac{(1+cos2x)(2sin2x) - (1-cos2x)(-2sin2x)}{(1+cos2x)^2} = \frac{4sin2x}{(1+cos2x)^2}$

That's good. Now, the proof. Expand the denominator;

$(1+cos2x)^2 = 1 + 2cos2x + cos^2(2x)$

You see that the answer include angles in x, not 2x? Well convert all the double angles to single ones.

$sin2x = 2cosx\,sinx$

$cos2x = 2cos^2 x - 1$

$cos^2 2x = (cos2x)^2 = 4cos^4 x - 4cos^2 x +1$

To summarise;

$\frac{4sin2x}{(1+cos^2 2x)^2} = \frac{4(2sinxcosx)}{1+2(2cos^2 x -1) + (4cos^4 x - 4cos^2 x + 1} = \frac{8sinxcosx}{4cos^4 x} = \frac{2sinx}{cos^3 x} = \frac{sinx}{cosx} . \frac{1}{cos^2 x}$

See it? :)

Sigh, my mouse stopped working... :(

thinay · 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.. :)

Unknown008 · Aug 13, 2009, 08:31 AM

Ok, for the last one then you did it right, just need to simplify.

$(1+\frac{dy}{dx})sec^2 (x+y) = sec^2(x+y) + \frac{dy}{dx}sec^2(x+y)$

Group the dy/dx:

$\frac{dy}{dx} = sec^2(x+y) + \frac{dy}{dx}sec^2(x+y)$

$\frac{dy}{dx} - \frac{dy}{dx}sec^2(x+y)= sec^2(x+y)$

Factorise dy/dx;

$\frac{dy}{dx}(1 - sec^2(x+y))= sec^2(x+y)$

$\frac{dy}{dx}(-tan^2(x+y))= sec^2(x+y)$

$\frac{dy}{dx}= \frac{sec^2(x+y)}{(-tan^2(x+y))}$

$\frac{dy}{dx}= \frac{\frac{1}{cos^2(x+y)}}{(-\frac{sin^2(x+y)}{cos^2(x+y)})}$

The cos^2(x+y) goes out;

$\frac{dy}{dx}= \frac{1}{-sin^2(x+y)}$

Which is:

$\frac{dy}{dx}= -csc^2(x+y)$

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

thinay · 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). :)

thinay · 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). :)

Unknown008 · Aug 13, 2009, 08:47 AM

Oh, I just figured it out! Oh my God!!

$-csc^2(x+y) = -cot^2(x+y) -1 = -\frac{1}{tan^2(x+y)} - 1 = -\frac{1}{y^2} - 1 = -(y^{-2} +1) = -(1 + y^{-2})$

Yay! LOL! :p

thinay · 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... :)

Unknown008 · Aug 13, 2009, 09:07 AM

Did you see earlier what y stands for?

Originally Posted by thinay

y = tan(x+y)

thinay · 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

thinay · 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 ((1-cos4x)/(1+cos4x))
the answer in the 2nd derivative will be: 8sec^2 2x |tan2x|

This one, its really giving me a headache.. :(

Unknown008 · Aug 13, 2009, 09:36 AM

OMG!! I don't know where the modulus came...

$y = \frac{(1-cos4x)^{\frac12}}{(1+cos4x)^{\frac12}}$

$\frac{dy}{dx} = \frac{(1+cos4x)^{\frac12} \frac12 (1-cos4x)^{-\frac12}.4sin4x - (1-cos4x)^{\frac12} \frac12(1+cos4x)^{-\frac12}.-4sin4x}{((1+cos4x)^\frac12)^2}$

Phew!!

$\frac{dy}{dx} = \frac{2(1+cos4x)^{-\frac12}(1-cos4x)^{-\frac12}[(1+cos4x)sin4x + (1-cos4x)sin4x]}{(1+cos4x)}$

$\frac{dy}{dx} = \frac{2(1+cos4x)^{-\frac12}(1-cos4x)^{-\frac12}(2sin4x)}{(1+cos4x)}$

Oh my... did you get that?

Unknown008 · Aug 13, 2009, 09:41 AM

Now;

$(1+cos4x)^{-\frac12}(1-cos4x)^{-\frac12} = \frac{1}{\sqrt{1+cos4x}\sqrt{1-cos4x}} = \frac{1}{\sqrt{(1+cos4x)(1-cos4x)}} = \frac{1}{\sqrt{1-cos^2 4x}} = \frac{1}{\sqrt{sin^2 4x}} = \frac{1}{sin 4x}$

$\frac{dy}{dx} = \frac{4 sin 4x}{(sin4x)(1+cos4x)}$

$\frac{dy}{dx} = \frac{4}{(1+cos4x)}$

Now, can you find the second derivative?

thinay · 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.. :))

thinay · 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

Unknown008 · 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