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    How do I differentiate this with respect to n??

     a^rb^{n-r}

    thanks for the help in advance.
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    1+1=2 , hope this helps
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    hilarious
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    \frac {d}{dn} (a^rb^{n-r}) = a^r(\frac {d}{dn} (b^{n-r})) = a^r(\frac {d}{dn} (\frac {b^n}{b^r}))

    i'm probably wrong but if i'm not that should be easier
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    (Original post by adie_raz)
    How do I differentiate this with respect to n??

     a^rb^{n-r}

    thanks for the help in advance.
    Is r a constant or a variable?
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    (Original post by Kolya)
    Is r a constant or a variable?
    It is variable. That is what stumped me.
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    (Original post by adie_raz)
    How do I differentiate this with respect to n??

     a^rb^{n-r}

    thanks for the help in advance.
     a^rb^{n-r}=a^r\frac{b^n}{b^r} a^r and b^r are constants...now say  y = a^rb^{n-r} , using what I have told you plus a bit of implicit differentiation, you should get your answer...oh and here's a hint to get you started; using natural logs (ln) is a good way to start
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    EDIT; if r is a variable then what I have said does not hold true...unless you want to partially differentiate it
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    (Original post by adie_raz)
    It is variable. That is what stumped me.
    So product rule to begin with, and then apply the quotient rule to b^(n-r).
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    (Original post by Kolya)
    So product rule to begin with, and then apply the quotient rule to b^(n-r).
    thank you
 
 
 

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