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    differentiate with respect to t: (β/(β-t))^α

    I get
    α*β^α / (β-t)^(α+1)

    Is that correct? The method used by the textbook is different: they've done (see image) whereas I got the (β-t))^α in the numerator and then used the chain rule Name:  diff.jpg
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    (Original post by zomgleh)
    differentiate with respect to t: (β/(β-t))^α

    I get
    α*β^α / (β-t)^(α+1)

    Is that correct? The method used by the textbook is different: they've done (see image) whereas I got the (β-t))^α in the numerator and then used the chain rule Name:  diff.jpg
Views: 51
Size:  2.9 KB
    They both work out the same, and they're both missing a minus sign.
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    (Original post by ghostwalker)
    they're both missing a minus sign.
    how come? I was almost sure the textbook solution would be correct?
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    (Original post by zomgleh)
    how come? I was almost sure the textbook solution would be correct?
    My mistake! Grovel, grovel!

    I was confusing the alpha exponent in the numerator, with the one in the denominator.
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    (Original post by ghostwalker)
    My mistake! Grovel, grovel!

    I was confusing the alpha exponent in the numerator, with the one in the denominator.
    haha phew! thanks anyway, you've helped me way too many times, appreciate it!
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    (Original post by ghostwalker)
    My mistake! Grovel, grovel!

    I was confusing the alpha exponent in the numerator, with the one in the denominator.
    Ok so I've done this one before but looking back at it I can't seem to figure it out-- dy/dx of y=(x-y)^3
    answer is 3(x-y)^2 / (1+ 3(x-y)^2
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    (Original post by zomgleh)
    Ok so I've done this one before but looking back at it I can't seem to figure it out-- dy/dx of y=(x-y)^3
    answer is 3(x-y)^2 / (1+ 3(x-y)^2
    Treat it with the chain rule. What do you get for your first line of working?
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    (Original post by ghostwalker)
    Treat it with the chain rule. What do you get for your first line of working?
    I get 3(x-y)^2
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    (Original post by zomgleh)
    I get 3(x-y)^2
    You had y=(x-y)^3

    So, using the chain rule we get y'=3(x-y)^2(1-y')

    Make sure you understand it, and then just rearrange.
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    (Original post by ghostwalker)
    You had y=(x-y)^3

    So, using the chain rule we get y'=3(x-y)^2(1-y')

    Make sure you understand it, and then just rearrange.
    got it, you have once again saved the day! voted you for the most helpful member btw, seeing as I probably won't be able to answer any of your doubts :P
 
 
 
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