Is this easy derivative question correct?

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Is this easy derivative question correct?

Mathematics
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You 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.

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Anyone have an idea? I can post the formula
Gotta love Patrick JMT on youtube
um um um
lol yes zepdrix? :)
\[\Large\rm f(x)=2+x^2+\tan\left(\frac{\pi}{2}x\right)\]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.
So then uhhhhh.. ya our formula stuff.
Maybe so, looks good?
Ooo 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 +_+
Its alright I understand this problem is a doozy
Oh oh oh, I'm being silly.. it's not f(0) in the bottom it's f'(0)
ya ya ya, that should fix things up :) find your derivative before plugging 0 in
\[\Large\rm (f^{-1})(2)=\frac{1}{f'(\color{orangered}{f^{-1}(2)})}\] \[\Large\rm (f^{-1})(2)=\frac{1}{f'(\color{orangered}{0})}\]
Eek whats the d dx?
Umm 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.
Ok so Im right then?
oh with the blue dot? ahhh yes, looks correct!! :)
THANK YOU!!!! WHooo hoooo!!!!
yay team \c:/

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