Rote Learner Posted February 3, 2008 Share Posted February 3, 2008 I am currently taking calc 2 because I am lazy and wanted to hold off taking calc 2 because calc one gave me some trouble. I am a mechanical engineering student and I need some help with this one problem. So far I got 18 out of the 19 questions correct, this one is just killing me and I think its just a silly mistake. Thank you Link to comment Share on other sites More sharing options...
Dave Posted February 4, 2008 Share Posted February 4, 2008 Just apply the quotient rule: [math]f(x) = \frac{g(x)}{h(x)} \Rightarrow f'(x) = \frac{h'(x)g(x) - g'(x)h(x)}{(h(x))^2}[/math] So set [imath]g(x) = x[/imath], [imath]h(x) = 1 - \ln(x-1)[/imath], work out their derivatives and go from there. Link to comment Share on other sites More sharing options...
Bignose Posted February 4, 2008 Share Posted February 4, 2008 Just apply the quotient rule: [math]f(x) = \frac{g(x)}{h(x)} \Rightarrow f'(x) = \frac{h'(x)g(x) - g(x)h'(x)}{(h(x))^2}[/math] So set [imath]g(x) = x[/imath], [imath]h(x) = 1 - \ln(x-1)[/imath], work out their derivatives and go from there. Small, but significant typo there Dave. The second term in the numerator should be g'h not gh' [math]f(x) = \frac{g(x)}{h(x)} \Rightarrow f'(x) = \frac{h'(x)g(x) - g'(x)h(x)}{(h(x))^2}[/math] Link to comment Share on other sites More sharing options...
Dave Posted February 6, 2008 Share Posted February 6, 2008 Quite right. Thanks for that. Link to comment Share on other sites More sharing options...
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