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

1. AmTran_Bus

2. AmTran_Bus

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.

3. AmTran_Bus

@welshfella

4. AmTran_Bus

Anyone have an idea? I can post the formula

5. AmTran_Bus

Gotta love Patrick JMT on youtube

6. zepdrix

um um um

7. AmTran_Bus

lol yes zepdrix? :)

8. 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.

9. zepdrix

So then uhhhhh.. ya our formula stuff.

10. AmTran_Bus

Maybe so, looks good?

11. zepdrix

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 +_+

12. AmTran_Bus

Its alright I understand this problem is a doozy

13. zepdrix

Oh oh oh, I'm being silly.. it's not f(0) in the bottom it's f'(0)

14. zepdrix

ya ya ya, that should fix things up :) find your derivative before plugging 0 in

15. zepdrix

$\Large\rm (f^{-1})(2)=\frac{1}{f'(\color{orangered}{f^{-1}(2)})}$ $\Large\rm (f^{-1})(2)=\frac{1}{f'(\color{orangered}{0})}$

16. AmTran_Bus

Eek whats the d dx?

17. zepdrix

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.

18. AmTran_Bus

Ok so Im right then?

19. zepdrix

oh with the blue dot? ahhh yes, looks correct!! :)

20. AmTran_Bus

THANK YOU!!!! WHooo hoooo!!!!

21. zepdrix

yay team \c:/