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AmTran_Bus
 one year ago
Is this easy derivative question correct?
AmTran_Bus
 one year ago
Is this easy derivative question correct?

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AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.2You don't have to walk me through, I know I need to be doing the work, but I would sure appreciate it if you would tell me if it is wrong so I can re do it.

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.2Anyone have an idea? I can post the formula

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.2Gotta love Patrick JMT on youtube

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.2lol yes zepdrix? :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large\rm f(x)=2+x^2+\tan\left(\frac{\pi}{2}x\right)\]So we want to evaluate this at the inverse functions x=2, which will correspond to the original functions y=2, ya? \[\Large\rm 2=2+x^2+\tan\left(\frac{\pi}{2}x\right)\]So uhhh.. it looks like this is leading to an x value of x=0.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So then uhhhhh.. ya our formula stuff.

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.2Maybe so, looks good?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Ooo ya I made a boo boo somewhere :P Cause f(0) is = 2. So we would end up with 1/2. Woops. Not an option. Thinkingggg +_+

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.2Its alright I understand this problem is a doozy

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Oh oh oh, I'm being silly.. it's not f(0) in the bottom it's f'(0)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2ya ya ya, that should fix things up :) find your derivative before plugging 0 in

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large\rm (f^{1})(2)=\frac{1}{f'(\color{orangered}{f^{1}(2)})}\] \[\Large\rm (f^{1})(2)=\frac{1}{f'(\color{orangered}{0})}\]

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.2Eek whats the d dx?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Umm I guess it would be \[\Large\rm f(x)=0+2x+\sec^2\left(\frac{\pi}{2}x\right)\cdot\frac{\pi}{2}\]I applied chain rule on that last term.

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.2Ok so Im right then?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2oh with the blue dot? ahhh yes, looks correct!! :)

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.2THANK YOU!!!! WHooo hoooo!!!!
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