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1+1=3?



 

 

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Guh?



Kimi wa ne tashika ni ano toki watashi no soba ni ita

Itsudatte itsudatte itsudatte

Sugu yoko de waratteita

Nakushitemo torimodosu kimi wo

I will never leave you

i'd rather not respond, but since a wrong answer has been given, i should probably correct that person.

but first, i imagine "modulus" is just a synonym for "absolute value", since "modulus" to me means the magnitude of a complex number, in which case the symbol "z" is more often used than "x".  i'll proceed assuming modulus means absolute value.

to approach this problem, first plot the function.  this is fairly easy, actually: you know how to plot f(x)=x, and you know how to plot f(x)=sin(x).  visually multiply the two, and you get f(x)=x sin(x).  going from sin(x) to sin(8 x) is easy, you just "squish" your graph 8 times in the horizontal direction.

to get the |sin(8 x)| part right, just remember that the |...| means you always end up with a positive value.  so for x |sin(8 x)|, which is the function x multiplied by sin(8 x), when x>0, this function is positive.  when x<0, this function is negative.

now you should know what the graph looks like.  it's basically a increasing sinusoidal with bumps.

it's easy to tell what points are not differentiable once you have the graph in front of you.  all the "flipping" points, i.e. n*Pi/8 for all n except n=0, are not differentiable.

the point n=0 is differentiable, with a derivative of 0.  you're probably not required to prove it, but you'll get extra credit if you prove it.  basically, prove that the left derivative and the same as the right derivative.  if you don't know what i'm saying you'll certainly not required to prove it.

now onto the actual derivative.  the |...| is a distraction, you should first get your "domains" right (you should know that domain is a fancy word for where your function is defined on). here goes:

|sin(8x)|  is:

            sin(8x)   when n*Pi/8 < x < (n+1)*Pi/8 for n = even  

           -sin(8x)   when n*Pi/8 < x < (n+1)*Pi/8 for n= odd    

 

so taking derivative of x sin(8x) or -x sin(8x) on their appropriate domains is simply

           sin(8x) + 8x cos(8x)  when n*Pi/8 < x < (n+1)*Pi/8 for n = even  

          -sin(8x) - 8x cos(8x)   when n*Pi/8 < x < (n+1)*Pi/8 for n= odd

           and is = 0 when x=0, and undefined for all x = n*Pi/8 except for n=0.

 



the Wii is an epidemic.

largedarryl said:
Okay, so now that I've refreshed my calc days.

df/dx = sin(8x) + 8x*cos(8x)/|cos(8x)| where cos(8x) != 0

 

the |cos(8x)| does not appear in the denominator.



the Wii is an epidemic.

Thanks for the refresher Lingyis, it's funny how quickly the basic calculus leaves your brain when you finish school, but I'm pretty sure my second attempt is the same as yours (only yours is more of a proof).

EDIT:  Are you sure |cos(8x)| does not appear in the denominator?  That is the only way  I can see the df/dx working out to a non-real value (divide by 0).



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Lingyis said:

i'd rather not respond, but since a wrong answer has been given, i should probably correct that person.

but first, i imagine "modulus" is just a synonym for "absolute value", since "modulus" to me means the magnitude of a complex number, in which case the symbol "z" is more often used than "x".  i'll proceed assuming modulus means absolute value.

to approach this problem, first plot the function.  this is fairly easy, actually: you know how to plot f(x)=x, and you know how to plot f(x)=sin(x).  visually multiply the two, and you get f(x)=x sin(x).  going from sin(x) to sin(8 x) is easy, you just "squish" your graph 8 times in the horizontal direction.

to get the |sin(8 x)| part right, just remember that the |...| means you always end up with a positive value.  so for x |sin(8 x)|, which is the function x multiplied by sin(8 x), when x>0, this function is positive.  when x<0, this function is negative.

now you should know what the graph looks like.  it's basically a increasing sinusoidal with bumps.

it's easy to tell what points are not differentiable once you have the graph in front of you.  all the "flipping" points, i.e. n*Pi/8 for all n except n=0, are not differentiable.

the point n=0 is differentiable, with a derivative of 0.  you're probably not required to prove it, but you'll get extra credit if you prove it.  basically, prove that the left derivative and the same as the right derivative.  if you don't know what i'm saying you'll certainly not required to prove it.

now onto the actual derivative.  the |...| is a distraction, you should first get your "domains" right (you should know that domain is a fancy word for where your function is defined on). here goes:

|sin(8x)|  is:

            sin(8x)   when n*Pi/8 < x < (n+1)*Pi/8 for n = even  

           -sin(8x)   when n*Pi/8 < x < (n+1)*Pi/8 for n= odd    

 

so taking derivative of x sin(8x) or -x sin(8x) on their appropriate domains is simply

           sin(8x) + 8x cos(8x)  when n*Pi/8 < x < (n+1)*Pi/8 for n = even  

          -sin(8x) - 8x cos(8x)   when n*Pi/8 < x < (n+1)*Pi/8 for n= odd

           and is = 0 when x=0, and undefined for all x = n*Pi/8 except for n=0.

 

Wasn't this my answer? at least I got the derivative right!

 



I have a hard time reading that stuff. I prefer seeing it written out, otherwise I can't do sqwat :(



4 ≈ One

lightbleeder said:
Lingyis said:

i'd rather not respond, but since a wrong answer has been given, i should probably correct that person.

but first, i imagine "modulus" is just a synonym for "absolute value", since "modulus" to me means the magnitude of a complex number, in which case the symbol "z" is more often used than "x".  i'll proceed assuming modulus means absolute value.

to approach this problem, first plot the function.  this is fairly easy, actually: you know how to plot f(x)=x, and you know how to plot f(x)=sin(x).  visually multiply the two, and you get f(x)=x sin(x).  going from sin(x) to sin(8 x) is easy, you just "squish" your graph 8 times in the horizontal direction.

to get the |sin(8 x)| part right, just remember that the |...| means you always end up with a positive value.  so for x |sin(8 x)|, which is the function x multiplied by sin(8 x), when x>0, this function is positive.  when x<0, this function is negative.

now you should know what the graph looks like.  it's basically a increasing sinusoidal with bumps.

it's easy to tell what points are not differentiable once you have the graph in front of you.  all the "flipping" points, i.e. n*Pi/8 for all n except n=0, are not differentiable.

the point n=0 is differentiable, with a derivative of 0.  you're probably not required to prove it, but you'll get extra credit if you prove it.  basically, prove that the left derivative and the same as the right derivative.  if you don't know what i'm saying you'll certainly not required to prove it.

now onto the actual derivative.  the |...| is a distraction, you should first get your "domains" right (you should know that domain is a fancy word for where your function is defined on). here goes:

|sin(8x)|  is:

            sin(8x)   when n*Pi/8 < x < (n+1)*Pi/8 for n = even  

           -sin(8x)   when n*Pi/8 < x < (n+1)*Pi/8 for n= odd    

 

so taking derivative of x sin(8x) or -x sin(8x) on their appropriate domains is simply

           sin(8x) + 8x cos(8x)  when n*Pi/8 < x < (n+1)*Pi/8 for n = even  

          -sin(8x) - 8x cos(8x)   when n*Pi/8 < x < (n+1)*Pi/8 for n= odd

           and is = 0 when x=0, and undefined for all x = n*Pi/8 except for n=0.

 

Wasn't this my answer? at least I got the derivative right!

 

That is the same as the one you listed, but I think he was putting this one down because my first response was really wrong.

 



thanks guys for all your help.

The computer programme I'm using is saying that all your answers for the derivative are wrong lol.

But it accepted npi/8 for where it wasn't defined. Thanks.

Sorry I didn't respond, I fell asleep haha.



To be honest I'm over calculus so I can't be bothered helping.

I am however interested by the notation everyone is using. Should the differential of f(x) be written as f'(x)? I thought dy/dx notation wasn't used when the original function was written as a function of a variable rather than a variable on its own.