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anonymous
 4 years ago
Differentiate sin(x) from first principles
anonymous
 4 years ago
Differentiate sin(x) from first principles

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hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2use sin Asin B formula, do u know it ?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2nopes, sin A  sin B on sin(x+h)  sin x.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I was thinking on starting out like this: \[\lim_{h \rightarrow 0}\frac{ \sin (x+h)  \sin(x) }{ h }\]

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2thats the correct start, i was giving a hint on how would u start simplifying numerator. using sin Asin B with A=x+h,B=x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0us, would you expand sin(x+h)?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2i think that would also work, there u have to separate out denominator's h to all numerator terms.....i personally find it easier to do with sin Asin B...and less lengthy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how does sinAsinB work?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2sin A  sin B = 2 cos (A+B)/2 sin (AB)/ 2 right ? so sin (x+h)sin x = 2 cos (x+h/2) sin (h/2) ok ? now put h=0 directly in cos term and use sin t/t = 1 in sin term...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry, don't know sinAsinB = 2 cos((A+B)/2)sin((A+B)/2)

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3It's... actually \[sinA  sinB = 2cos\frac{A+B}{2}sin\frac{AB}{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Q: Is that the product to sums formula?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2then u better go with sin(x+h) = sin x cos h + cos x sin h ....

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3sumtoproduct indeed.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lim... [ sin(x)cos(h)+cos(x)sin(h)sin(x) ] / h lim... [ sin(x)(cos(h)1)+cos(x)sin(h) ] / h then what?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2can u solve lim x>0 (1cos x)/x = 1/2 ?? pretty standard. the other limit is pretty easy, just use sin h/h =1

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3Hmmm... Isn't it lim x>0 (1cos x)/x = 0?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2made a mistake....wait.yup its 0,sorry.

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2but basically u need to separate the numerator in 2 limits.

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2so your 1st limit will be 0 as pointed out correctly and 2nd limit will be just cos x *1 = cos x

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3\[\lim_{h \rightarrow 0}\frac{sin(x)(cos(h)1)+cos(x)sin(h) }{h}\]\[=\lim_{h \rightarrow 0}(\frac{sin(x)(cos(h)1)}{h}\ \ +\frac{cos(x)sin(h) }{h})\]\[=\lim_{h \rightarrow 0}sin(x)\frac{(cos(h)1)}{h} \ +\lim_{h \rightarrow 0}cos(x)\frac{sin(h) }{h}\]\[=...\]

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3Perhaps one more step for you... \[=sin(x)\lim_{h \rightarrow 0}\frac{(cos(h)1)}{h} \ +cos(x) \lim_{h \rightarrow 0}\frac{sin(h) }{h}\] I think you can evaluate the limit now..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how are theses limits calculated?

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3As @hartnn mentioned, use these: \[\lim_{x \rightarrow 0}\frac{1cos(x)}{x} =0\]\[\lim_{x \rightarrow 0}\frac{sin(x)}{x} =1\]

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2actually the 1st one isn't standard, u need to solve it by multiplying and dividing by (1+cos x) which gives sin^2 x in numerator and using 2nd limit to get answer as 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah um, how are these proven? (Sorry if I'm being a bit fussy)

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3@apple_pi Do you have time to watch a video for explanations?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe later... just send me the links

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2there are many ways to prove lim sin x/x =1 the simplest one is the use of L'Hospitals rule, but since u are starting limits, i guess u don't know about this rule..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh yeah, MIT OCW is good. Does this show the whole d/dx sin(x)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks...turns out I downloaded this ages ago, but stopped after lecture 2 and jumped about places

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3Here it goes. The notes I've copied from that lecture. IMG.pdf should contain all three pages.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks a heap. Can you show me how sinAsinB = ...

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3Do you know how to work out the producttosum formula?

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3Example: sinAcosB = (1/2) [ sin(A+B) +sin(AB)]

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2that u can get using sin (xy) and sin (x+y) formula, u know that right ? just solve sin(x+y)  sin (xy) = ....... and then put x+y = A xy = B so x= (A+B)/2 y=(AB)/2.

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.3hartnn explained, I think :)

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2so whats sin(x+y)  sin (xy)= ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sin(x)cos(y)+cos(x)sin(y)sin(x)cos(y)+cos(x)sin(y) = 2cos(x)sin(y)

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2thats correct, now just put values of x and y as i mentioned, u understood, how i found x and y in terms of A and b .. ?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2x+y=A xy=B Adding x+x=A+B >x=(A+B)/2 i think y can can find now...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(A+B)/2+y=A y=(AB)/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0uh, so does that mean sin(A)sin(B)=2cos((A+B)/2)sin((AB)/2)

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2similarly u can find sin A+sin B cos A + cos B cos A  cos B and remember them as they are standard formulas.
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