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I needed some help finding the derivative of the following at $m=15$ $$r = \left(\frac{m^2}{a}\right)\left(\frac{1}{b}-\frac{m}{c}\right)$$

Where $a,b,c$ are constants

I have no idea where to start. Can someone please point me in the correct direction?

Thanks!

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  • $\begingroup$ use $(f(m)g(m))'=f'(m)g(m)+f(m)g'(m)$ $\endgroup$ – user60610 Feb 21 '13 at 19:18
  • $\begingroup$ @TongZhang I looked at that but how would you find the derivative of the inside of the second portion? $\endgroup$ – Jeel Shah Feb 21 '13 at 19:20
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$$r =\frac{m^2}{a}\left(\frac{1}{b}-\frac{m}{c}\right)$$ $$r' =\frac{2m}{a}\left(\frac{1}{b}-\frac{m}{c}\right)+\frac{m^2}{a}\left(-\frac{1}{c}\right)$$ $$r'(15) =\frac{2\cdot 15}{a}\left(\frac{1}{b}-\frac{15}{c}\right)+\frac{15^2}{a}\left(-\frac{1}{c}\right)$$ $$r'(15) =\frac{30}{a}\left(\frac{1}{b}-\frac{15}{c}\right)+\frac{225}{a}\left(-\frac{1}{c}\right)$$

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  • $\begingroup$ Hi Adi, miss you at FB. $\endgroup$ – mrs Feb 21 '13 at 19:24

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