Help me to solve these questions.. Mathematics T 1

[uchiha_itachi]

Hi all,i need yours help to solve my unsolved questions on math t 1.. :P :):):)

Diffrentiate each of the following functions with respect to x. [use chain rule]
1.) ln(sin x cos x)

More to come.. hehehe

[luke]

are you asking us to do your homework or something?

[jiacheng91]

is it (cos2x)/(sinxcosx):))

[uchiha_itachi]

are you asking us to do your homework or something?
nope.. but if you dont mind, you can post your solution for my question..
and then, we all can disscuss it..
i've got stuck to solve it.. thanks

sorry 4 my bad language..

[castle]

ln (sinx cosx)=ln sinx +ln cosx
=cosx/sinx +(-sinx/cosx)
=cotx-tanx

[uchiha_itachi]

ln (sinx cosx)=ln sinx +ln cosx
=cosx/sinx +(-sinx/cosx)
=cotx-tanx
thanks.. got idea at your steps but its not final answer actually..
actually, i have the answer but not the solution.. :amuse

sooo soorry .. im forgot to tell you all to use chain rule.. edited..

[castle]

what is the answer?mathematical answers are like speech,they can be in different forms but convey the same meaning...

[uchiha_itachi]

what is the answer?mathematical answers are like speech,they can be in different forms but convey the same meaning...
the answer is 2 cot 2x... :cry ive got headhache to think the right solution..http://www.recom.org/forum/images/icons/icon9.gif

My steps :

y = ln u so, dy/du = 1/u
u = sinx cos x so, du/dx = -sinx cos x ???

then, im stuck at this step..

dy/dx = -sinx cosx / sinx cos x next step????? or my step is wrong..

[nicodemus]

y = ln (sin x cos x)

dy/dx = [1/(sin x cos x)] [d/dx (sin x cos x)]

= [1/(sin x cos x)] [(sin x)(- sin x) + (cos x)(cos x)]

= [1/(sin x cos x)] [cos2 x - sin2 x]

= cos 2x / (sin x cos x)

= cos 2x (2/sin 2x)

= 2 cot 2x

QED...

btw.. What is chain rule? Is it first principle? Never heard of it before la during my form 6... =P

[uchiha_itachi]

y = ln (sin x cos x)

dy/dx = [1/(sin x cos x)] [d/dx (sin x cos x)]

= [1/(sin x cos x)] [(sin x)(- sin x) + (cos x)(cos x)]

= [1/(sin x cos x)] [cos2 x - sin2 x]

= cos 2x / (sin x cos x)

= cos 2x (2/sin 2x)

= 2 cot 2x

QED...

btw.. What is chain rule? Is it first principle? Never heard of it before la during my form 6... =P


Nice one bro.. http://www.recom.org/forum/images/icons/icon10.gif where are u study now?
chain rule? i thought maybe it is the new rule to make it look easy..http://www.recom.org/forum/images/icons/icon10.gif
i take it at pelangi book..

[lawteoh]

Chain rule is dy/du x du/dx i think.. wow, nicodemus, u still remember form 6 maths? great.. =)

[uchiha_itachi]

Chain rule is dy/du x du/dx i think.. wow, nicodemus, u still remember form 6 maths? great.. =)
yes. u are right.. http://www.recom.org/forum/images/icons/icon10.gif dy/du x du/dx = chain rule ..

then, the next question i think.. :p

[nicodemus]

Nice one bro.. http://www.recom.org/forum/images/icons/icon10.gif where are u study now?
chain rule? i thought maybe it is the new rule to make look it easy..http://www.recom.org/forum/images/icons/icon10.gif
i take it at pelangi book..

USM medicine... I see...

Chain rule is dy/du x du/dx i think.. wow, nicodemus, u still remember form 6 maths? great.. =)

Really? But that rule is used for composite functions only, never heard of it being called as chain rule... What ever, doesn't concerns us anyway... =D

=D thanks... I try to remember it, to keep my brain from being rusty and filled with all those medical stuff, they're boring... I miss doing maths actually... ;-)

Next ques? Bring it on... =P

[uchiha_itachi]

ok then,

if y = 1/[(x-1)^1/2]sinx , find dy/dx when x=n/2 .. n is pai... got it? hehehe

[nicodemus]

This is a bit difficult... Is the answer 1.1594?

And to confirm, in this equation y = 1/[(x-1)^1/2]sinx, the denominator is
[(x-1)^0.5][sin x], am i right?

[zekrypton]

Er.... sinxcosx bukan (sin2x)/2 ke?

[uchiha_itachi]

This is a bit difficult... Is the answer 1.1594?

And to confirm, in this equation y = 1/[(x-1)^1/2]sinx, the denominator is
[(x-1)^0.5][sin x], am i right?

nice try bro.. but it is not the correct answer ..

the answer is -[1/[2(1/2pai-1)^3/2]] hehehehe

now, i think you can give me the correct solution... http://www.recom.org/forum/images/icons/icon10.gif

Er.... sinxcosx bukan (sin2x)/2 ke?
sin(2a) = 2(cos a) (sin a) ye xye jugak.. hehehe

[nicodemus]

nice try bro.. but it is not the correct answer ..

the answer is -[1/[2(1/2pai-1)^3/2]] hehehehe

now, i think you can give me the correct solution... http://www.recom.org/forum/images/icons/icon10.gif

My answer is correct, if you bother to convert the answer given to a value...

Well, you have 2 methods... Either do the multiplication method or division method... I'll show you the division one as it's easier...

if you differentiate y = u/v, dy/dx would be [v(du/dx)-u(dv/dx)] / v^2

So you change your question to this form which is y = [(x-1)^(-1/2)] / sin x

dy/dx = < (sin x)[-1/2(x-1)^(-3/2)] - [((x-1)^(-1/2))(cos x)] > / sin^2 x

separate them to be...

= -1/[2 sin x (x-1)^(3/2)] - cot x / [sin x (x-1)^(1/2)]

when x = pi/2, sin x = 1, cot x = 0, so you'll get

= -1/[2(pi/2 -1)^(3/2)]

Done.... :-)

[uchiha_itachi]

My answer is correct, if you bother to convert the answer given to a value...

Well, you have 2 methods... Either do the multiplication method or division method... I'll show you the division one as it's easier...

if you differentiate y = u/v, dy/dx would be [v(du/dx)-u(dv/dx)] / v^2

So you change your question to this form which is y = [(x-1)^(-1/2)] / sin x

dy/dx = < (sin x)[-1/2(x-1)^(-3/2)] - [((x-1)^(-1/2))(cos x)] > / sin^2 x

separate them to be...

= -1/[2 sin x (x-1)^(3/2)] - cot x / [sin x (x-1)^(1/2)]

when x = pi/2, sin x = 1, cot x = 0, so you'll get

= -1/[2(pi/2 -1)^(3/2)]

Done.... :-)

my bad.. pardon me... hihihi http://www.recom.org/forum/images/icons/icon10.gif
i'll checkout your answer soon when i reach this question.. :P
now, i need to understand the easy question before it.. :p

oh, could you pls check your answer of my question before? is it right? becoz sin(2a) = 2(cos a) (sin a)

anyway, thanks

[nicodemus]

my bad.. pardon me... hihihi http://www.recom.org/forum/images/icons/icon10.gif
i'll checkout your answer soon when i reach this question.. :P
oh, could you pls check your answer of my question before?

anyway, thanks

Nvm... You are most welcome... ;-)

[uchiha_itachi]

y = ln (sin x cos x)

dy/dx = [1/(sin x cos x)] [d/dx (sin x cos x)]

= [1/(sin x cos x)] [(sin x)(- sin x) + (cos x)(cos x)]

= [1/(sin x cos x)] [cos2 x - sin2 x]

= cos 2x / (sin x cos x)

= cos 2x (2/sin 2x)

= 2 cot 2x

QED...

btw.. What is chain rule? Is it first principle? Never heard of it before la during my form 6... =P

Er.... sinxcosx bukan (sin2x)/2 ke?

Look like the answer in pelangi book is wrong.......

Edited... no, im wrong.. look like post zekrypton play important role in this case.. hahahha

[lawteoh]

USM medicine... I see...



Really? But that rule is used for composite functions only, never heard of it being called as chain rule... What ever, doesn't concerns us anyway... =D

=D thanks... I try to remember it, to keep my brain from being rusty and filled with all those medical stuff, they're boring... I miss doing maths actually... ;-)

Next ques? Bring it on... =P

Haha.. and to think im doing engineering course later.. oh well, ur maths is still very strong.. admire u for that when u hv not touch maths for so long.. =)

[uchiha_itachi]

Haha.. and to think im doing engineering course later.. oh well, ur maths is still very strong.. admire u for that when u hv not touch maths for so long.. =)
Agreed with lawteoh.. keep it up nicodemus and stay alive with this thread to help me ... :p

[Sillyboy]

Agreed with lawteoh.. keep it up nicodemus and stay alive with this thread to help me ... :p

Why are you posting your homework questions?

[uchiha_itachi]

Why are you posting your homework questions?
no.. its not my homework. it's my personal exercise actually.. :)

[nicodemus]

Haha.. and to think im doing engineering course later.. oh well, ur maths is still very strong.. admire u for that when u hv not touch maths for so long.. =)

Thanks lawteoh... :-D

Agreed with lawteoh.. keep it up nicodemus and stay alive with this thread to help me ... :p

Haha... Sure, I'm willing to help out... :-)

[uchiha_itachi]

Next tough question... (4 me now) hehehe

Diffrentiate this.. (chain rule)

(2x-3)^2/(3x+4)^3

[youngyew]

Actually uchiha_itachi, if you want a quick answer complete with the working steps, just head off to Wolfram Alpha and type in your question.

For your question:
http://www.wolframalpha.com/input/?i=differentiate+(2x-3)^2%2F(3x%2B4)^3 (http://www.wolframalpha.com/input/?i=differentiate+%282x-3%29%5E2%2F%283x%2B4%29%5E3)

[uchiha_itachi]

Actually uchiha_itachi, if you want a quick answer complete with the working steps, just head off to Wolfram Alpha and type in your question.

For your question:
http://www.wolframalpha.com/input/?i=differentiate+(2x-3)^2%2F(3x%2B4)^3 (http://www.wolframalpha.com/input/?i=differentiate+%282x-3%29%5E2%2F%283x%2B4%29%5E3)

amazing.. this method is too quick.. thanks..http://www.recom.org/forum/images/icons/icon10.gif
but a little bit difficult to understand.. hahaha

[nicodemus]

Amazing... I never knew there is such a website...

[uchiha_itachi]

Amazing... I never knew there is such a website...
So, now.. who is quick? nicodemus or wolframalpha? hahaha

[zekrypton]

Look like the answer in pelangi book is wrong.......

Edited... no, im wrong.. look like post zekrypton play important role in this case.. hahahha

Wah, i feel so honoured la woi. XDXDXD

Concerning that question, it's a normal differentiation question i think. Use quotient rule.

(denominator).d(numerator)/dx - (numerator).d(denominator)/dx
________________________________________________________
(denominator)^2

long time didn't touch maths adee.

[Sillyboy]

no.. its not my homework. it's my personal exercise actually.. :)

Don't rely on us to help you out all the time! Try to obtain the derivatives yourself! They are nothing but tedium and utilises the very basics of calculus:

Chain Rule (some people refer as the function of a function rule)
Quotient Rule

Good luck!

[uchiha_itachi]

(1 - y/x) / (1/y + lnx) = (xy-y^2) / (x + xy lnx)

But, my confusion is....
How we got that answer (xy-y^2) / (x + xy lnx)
i have tried in many ways, but still can't find how to got them... thanks.. :p

[jiacheng91]

(1-y/x)/(1/y+ln x)
= (x-y)/(x(1/y+ ln x)) take 1/x out from the top thingy, which makes the x go down thr

= (y(x-y))/x(1+ y ln x) take 1/y out from the btm thingy, which makes the y go up thr

= (xy-y^2)/(x+xy ln x)