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How can I find the derivative with respect to $v$ for the following two equations: $$[1-F(v)]^{n} \tag{1}$$ and $$1-[1-F(v)]^{n} \tag{2}$$ where $$F'(v)=f(v)$$

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  • $\begingroup$ Have you studied the chain rule? $\endgroup$ – David Mitra Feb 16 '12 at 22:26
  • $\begingroup$ Well, you have to show that by induction. $\endgroup$ – checkmath Feb 16 '12 at 22:41
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    $\begingroup$ Note: An "equation" must have an "equal" sign in it (hence the name). What you give are not "equations", they are expressions. $\endgroup$ – Arturo Magidin Feb 16 '12 at 22:43
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By using the chain rule: $$\Big((1-F(v))^n\Big)'=(1-F(v))'n(1-F(v))^{n-1}=-f(v)n(1-F(v))^{n-1}$$ Similarly: $$\Big(1-(1-F(v))^n\Big)'=0-(-f(v)n(1-F(v))^{n-1})=f(v)n(1-F(v))^{n-1}$$

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  • $\begingroup$ got it, thank you so much! $\endgroup$ – user1061210 Feb 16 '12 at 22:43

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