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September 25, 2025, 07:04:54 pm
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Author Topic: TyErd's questions  (Read 42418 times)  Share 

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TyErd

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Re: TyErd's questions
« Reply #210 on: May 30, 2010, 07:03:39 pm »
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Differentiate:
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m@tty

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Re: TyErd's questions
« Reply #211 on: May 30, 2010, 07:04:29 pm »
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.
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TyErd

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Re: TyErd's questions
« Reply #212 on: May 30, 2010, 07:29:36 pm »
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oh yeah so the derivative of that would be cos x
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m@tty

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Re: TyErd's questions
« Reply #213 on: May 30, 2010, 08:03:29 pm »
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Yep.
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TyErd

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Re: TyErd's questions
« Reply #214 on: June 09, 2010, 11:03:45 am »
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Differentiate:    and also
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the.watchman

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Re: TyErd's questions
« Reply #215 on: June 09, 2010, 11:09:13 am »
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Quote from: TyErd on June 09, 2010, 11:03:45 am
Differentiate:    and also

Well the first function can be written as:

,   
,   

So differentiate each part (noting that the cusp points are not differentiable)

You could also split the second function into parts to differentiate
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TyErd

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Re: TyErd's questions
« Reply #216 on: June 09, 2010, 11:29:12 am »
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Quote from: the.watchman on June 09, 2010, 11:09:13 am
Quote from: TyErd on June 09, 2010, 11:03:45 am
Differentiate:    and also
Well the first function can be written as:

,   


isn't the domain ?
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the.watchman

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Re: TyErd's questions
« Reply #217 on: June 09, 2010, 11:32:12 am »
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Well no, the modulus splits it up so that:

First part - , when , so when

Second part - , when , so when

Oh, and for the function itself, it is defined when x=0 and x=4, it's only the derived function which does not exist at these points :)
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TyErd

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Re: TyErd's questions
« Reply #218 on: June 09, 2010, 11:47:14 am »
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Quote from: the.watchman on June 09, 2010, 11:32:12 am
Well no, the modulus splits it up so that:

First part - , when , so when


okay, graphically I get it but algebraically I get
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the.watchman

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Re: TyErd's questions
« Reply #219 on: June 09, 2010, 12:03:36 pm »
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Be careful when doing it algebraically, this is the proper method:





CASE 1:

AND (two positives make a positive)

From these, (finding the intersection)

CASE 2:

AND (two negatives make a positive)

From these, (finding the intersection)

So the solutions are or :)
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TyErd

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Re: TyErd's questions
« Reply #220 on: June 11, 2010, 12:21:05 am »
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If , prove that .  I dont even understand what the expression is askin for.
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Re: TyErd's questions
« Reply #221 on: June 11, 2010, 12:30:11 am »
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Quote from: TyErd on June 11, 2010, 12:21:05 am
If , prove that .  I dont even understand what the expression is askin for.
its prettty simple i think:    dy/dx=4x^3
x*dy/dx= x*(4x^3)=4x^4=4(x^4) =4y=rhs
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TyErd

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Re: TyErd's questions
« Reply #222 on: June 11, 2010, 12:42:38 am »
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ohhh I getcha thanks!
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kenhung123

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Re: TyErd's questions
« Reply #223 on: June 11, 2010, 07:38:10 am »
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Hmm yea, the book doesn't really explain how to differentiate absolute value functions well...
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Juddinator

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Re: TyErd's questions
« Reply #224 on: June 13, 2010, 12:20:54 pm »
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P and Q lie on the curve . The x-co-ords of P and Q and 2 and (2+h) respectively. What is the gradient of PQ?

I tried to derive the equation of the curve ten just simply sub in the value of the x-co-ords but I got stuck when I realised you have2 x values, not one.
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