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I'm having trouble with 1F7.b) if anyone would be kind enough to help, it would be greatly appreciated. It seems to me like obvious choice is the quotient rule and chain rule to solve this, or possibly the product rule with the chain rule. If these are indeed the ways to approach the problem, I may need a step by step :(
Link to problem set: http://ocw.mit.edu/courses/mathematics/1801singlevariablecalculusfall2006/readings/e_exrcs_scsn_1_7.pdf
Link to solution: http://ocw.mit.edu/courses/mathematics/1801singlevariablecalculusfall2006/readings/s_solutns_exrcis.pdf
 one year ago
 one year ago
I'm having trouble with 1F7.b) if anyone would be kind enough to help, it would be greatly appreciated. It seems to me like obvious choice is the quotient rule and chain rule to solve this, or possibly the product rule with the chain rule. If these are indeed the ways to approach the problem, I may need a step by step :( Link to problem set: http://ocw.mit.edu/courses/mathematics/1801singlevariablecalculusfall2006/readings/e_exrcs_scsn_1_7.pdf Link to solution: http://ocw.mit.edu/courses/mathematics/1801singlevariablecalculusfall2006/readings/s_solutns_exrcis.pdf
 one year ago
 one year ago

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WaynexBest ResponseYou've already chosen the best response.1
Chain rule and quotient rule are what is needed here. Have you tried rewriting the radical in exponential form? You could do without the quotient rule if you make the exponential form of that radical a negative (i.e., the reciprocal form). It would go something like this:\[m _{0}(1\frac{ v^2 }{ c^2 })^\frac{ 1 }{ 2 }.\] We could make things visually simpler by realizing that v squared over c squared is this:\[(\frac{ v }{ c })^2,\]and then we can make a quick variable substitution of:\[u^2=(\frac{ v }{ c })^2.\] The substitution is tricky, so I suggest just trying it so that you can go through the motions and get an idea of where this one goes. Then, when you have that, if your answer doesn't match up to the solution, go back and do it without that u substitution.
 one year ago

BrentLBest ResponseYou've already chosen the best response.0
I got it with the substitution after a couple of attempts, thank you.
 one year ago
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