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C3 Differentiation Urgent Help needed!!!

Hi the question is:

The curve with equation y=(sin 1/2x)/2+cos 1/2x is defined for 0<x<2pi.

Show that dy/dx=(1/2 + cos 1/2x)/(2+cos 1/2x)^2

I've used the quotient rule and ended up with
dy/dx=(2+cos 1/2x((cos 1/2x)/2)-sin 1/2x((-sin 1/2x)/2))/(2+cos 1/2x)^2
and not sure where to go from there. Would really appreciate some help please, thanks :smile:

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Reply 1
bump!
Original post by jordanwu
bump!


Is the function y=sinx22+cosx2 y = \frac{ \sin \frac{x}{2}}{2 + \cos \frac{x}{2}} ? Typed maths does not read very well :-)
Reply 3
Original post by 16Characters....
Is the function y=sinx22+cosx2 y = \frac{ \sin \frac{x}{2}}{2 + \cos \frac{x}{2}} ? Typed maths does not read very well :-)


Yes it is :smile:
Original post by jordanwu
Hi the question is:

The curve with equation y=(sin 1/2x)/2+cos 1/2x is defined for 0<x<2pi.

Show that dy/dx=(1/2 + cos 1/2x)/(2+cos 1/2x)^2

I've used the quotient rule and ended up with
dy/dx=(2+cos 1/2x((cos 1/2x)/2)-sin 1/2x((-sin 1/2x)/2))/(2+cos 1/2x)^2
and not sure where to go from there. Would really appreciate some help please, thanks :smile:


Write out the numerator again, then multiply out the sin and cos and apply two identities to the resulting terms.
Original post by jordanwu
Yes it is :smile:


Then yeah you have applied the quotient rule correctly and should do what SeanFM said.
Reply 6
Original post by SeanFM
Write out the numerator again, then multiply out the sin and cos and apply two identities to the resulting terms.


I'm not sure what cos 1/2x multiplied by cos 1/2x is...
Reply 7
Anyone?
Original post by jordanwu
Anyone?


cos squared of x/2 (sorry I don't know the latex for cos squared)
Original post by jordanwu
I'm not sure what cos 1/2x multiplied by cos 1/2x is...


Original post by 16Characters....
Then yeah you have applied the quotient rule correctly and should do what SeanFM said.


Woops! 16Characters was kindly helping you already so I shall leave this one to him.
Reply 10
Original post by jordanwu
I'm not sure what cos 1/2x multiplied by cos 1/2x is...


cos2(x2)cos^2(\frac{x}{2})

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Original post by SeanFM
Woops! 16Characters was kindly helping you already so I shall leave this one to him.


Well strictly speaking you were the first to offer any actual help, I was just verifying the question :-)
Reply 12
Original post by 16Characters....
cos squared of x/2 (sorry I don't know the latex for cos squared)


If you wanna know the LaTeX look at my post
Reply 13
Original post by Andy98
cos2(x2)cos^2(\frac{x}{2})

Posted from TSR Mobile


So at the moment I have for the numerator:

cos (x/2) + (cos^2 (x/2) + sin^2 (x/2))/2

and I know that sin^2x + cos^2x = 1,

but I have sin^2 (x/2) and cos^2 (x/2) instead so what do I do?
Original post by Andy98
cos2(x2)cos^2(\frac{x}{2})

Posted from TSR Mobile


Damn that was simpler LaTeX than I had expected...
Original post by jordanwu
So at the moment I have for the numerator:

cos (x/2) + (cos^2 (x/2) + sin^2 (x/2))/2

and I know that sin^2x + cos^2x = 1,

but I have sin^2 (x/2) and cos^2 (x/2) instead so what do I do?


You can apply the normal identity. All that sin2α+cos2α=1 sin^2 \alpha + cos^2 \alpha = 1 means is that the sum of the squares of the sine and cosine of a given angle α \alpha is 1. x/2 is still an angle so the identity holds for that also.
Reply 16
Original post by 16Characters....
Damn that was simpler LaTeX than I had expected...


Hahaha yeah - I usually type what seems sensible and see what happens :lol:

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Reply 17
Original post by jordanwu
So at the moment I have for the numerator:

cos (x/2) + (cos^2 (x/2) + sin^2 (x/2))/2

and I know that sin^2x + cos^2x = 1,

but I have sin^2 (x/2) and cos^2 (x/2) instead so what do I do?


Carry on as normal, the identity works as long as the angles are the same.

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Reply 18
Ok thanks guys I got to the answer.
So the next part is: find the value of x at which the gradient of the curve is 1/12, giving your answer to 3sf.
So 1/12=(1/2 + cos (x/2)/(2 + cos (x/2))^2 , but I keep getting stuck on rearranging. Any tips?
Reply 19
bump!

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