anonymous
  • anonymous
f(x) = 3 + x^2 + tan (pix/2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Any one can help me with this derivative of an inverse
sweetburger
  • sweetburger
wait so your looking to take the derivative of this equation or what exactly?
anonymous
  • anonymous
yes Im in the last part but I dont get quite the exact answer

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anonymous
  • anonymous
I get (pi sec^2 pix/2)/(2) + 2x
freckles
  • freckles
why do you say derivative of inverse ?
freckles
  • freckles
Do you mean just find derivative?
sweetburger
  • sweetburger
so 3 gets deleted x^2 turns into 2x and tan(pix/2) = sec^2x(pix/2) times the derivative of pix/2 which i believe equates to pi/2. I am not sure if this is totally correct so ya check with someone else.
sweetburger
  • sweetburger
so it would look like 2x+sec^2x(pix/2)(pi/2) I think
anonymous
  • anonymous
yes, i evaluated at 0 but i get pi^2/4
sweetburger
  • sweetburger
x+sec^2(pix/2)(pi/2) is the correct value I accidentally left in x above.
freckles
  • freckles
I'm still confused why you said inverse
freckles
  • freckles
are you wanting to find \[(f^{-1})'(a)\] for some number a
anonymous
  • anonymous
yes freckles
freckles
  • freckles
what is the a
anonymous
  • anonymous
I used a couple of engines and I get (pi*sec^2(pix/2)(2))/ + 2x
anonymous
  • anonymous
a = 3
sweetburger
  • sweetburger
@Eco1 you you are right I made a typo and left out the 2 in front of the x
anonymous
  • anonymous
So I evaluate what I wrote above at zero but I dont get pi/2, which is the right asnwer
freckles
  • freckles
\[(f^{-1})'(3)=\frac{1}{f'(f^{-1}(3))} \\ f^{-1}(3)=? \implies f(?)=3 \\ \text{ so we need \to find ? such that }3=3+?^2+\tan(\frac{\pi ?}{2}) \\ \text{ and yes } f'(x)=0+2x+\frac{\pi}{2} \sec^2(\frac{\pi}{2}x)\]
freckles
  • freckles
that should be pretty easy to find ?
freckles
  • freckles
because we know both tan(0)=0 and 0^2=0
freckles
  • freckles
so ?=0
freckles
  • freckles
\[(f^{-1})'(3)=\frac{1}{f'(0)}\]
freckles
  • freckles
last thing for you to do is just to plug in 0 into your f' you found
anonymous
  • anonymous
I did multiple times but I dont get the right answer
freckles
  • freckles
\[f'(0)=2(0)+\frac{\pi}{2}[\sec(\frac{\pi}{2}(0))]^2\]
freckles
  • freckles
well first the inside of that sec( ) thing you can be simplified pi/2*0=0 so sec(0)=?
freckles
  • freckles
\[(f^{-1})'(3)=\frac{1}{f'(0)}=\frac{1}{2(0)+\frac{\pi}{2}[\sec(0)]^2}\]
freckles
  • freckles
you know cos(0)=1 so sec(0)=?
freckles
  • freckles
hint sec and cos are reciprocal functions of each other
anonymous
  • anonymous
reciprocal is synonym of inverse?
freckles
  • freckles
it depends what inverse you are talking about multiplicative inverse yeah
freckles
  • freckles
like sec(x)=1/cos(x) so sec(0)=1/cos(0)
freckles
  • freckles
1/cos(x) is reciprocal of cos(x)
freckles
  • freckles
some people just like to call it the flipping of
anonymous
  • anonymous
but it is sec(0)^2 not sec(0)
anonymous
  • anonymous
I meant sec^2(0)
freckles
  • freckles
if cos(0)=1 then sec(0)=1/cos(0)=1/1=1 if you square both sides of cos(0)=1 you get still cos^2(0)=1 so sec^2(0)=1 \[(f^{-1})'(3)=\frac{1}{f'(0)}=\frac{1}{0+\frac{\pi}{2}[1]^2}=\frac{1}{\frac{\pi}{2}}\]
freckles
  • freckles
1/(a/b) means just flip the number in the denominator
freckles
  • freckles
so 1/(a/b) is b/a
freckles
  • freckles
are you cool?
freckles
  • freckles
still processing?
anonymous
  • anonymous
Thanks man, or girl?
anonymous
  • anonymous
I just made an algebra error
anonymous
  • anonymous
I got it now, but this problem is a pain in the behind
freckles
  • freckles
I might be a she. And also it gets way more fun with more practice.
anonymous
  • anonymous
I think my algebra is rusty
freckles
  • freckles
it seems like everyone is working out their algebra kinks in calculus
freckles
  • freckles
so no worries
anonymous
  • anonymous
tell me about it
anonymous
  • anonymous
Any one out there who can give me a hand doing integrals
freckles
  • freckles
post a new question so people can see it :)

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