f(x) = 3 + x^2 + tan (pix/2)

- anonymous

f(x) = 3 + x^2 + tan (pix/2)

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- anonymous

Any one can help me with this derivative of an inverse

- sweetburger

wait so your looking to take the derivative of this equation or what exactly?

- anonymous

yes
Im in the last part but I dont get quite the exact answer

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## More answers

- anonymous

I get (pi sec^2 pix/2)/(2) + 2x

- freckles

why do you say derivative of inverse ?

- freckles

Do you mean just find derivative?

- 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

so it would look like 2x+sec^2x(pix/2)(pi/2) I think

- anonymous

yes, i evaluated at 0 but i get pi^2/4

- sweetburger

x+sec^2(pix/2)(pi/2) is the correct value I accidentally left in x above.

- freckles

I'm still confused why you said inverse

- freckles

are you wanting to find \[(f^{-1})'(a)\]
for some number a

- anonymous

yes freckles

- freckles

what is the a

- anonymous

I used a couple of engines and I get (pi*sec^2(pix/2)(2))/ + 2x

- anonymous

a = 3

- sweetburger

@Eco1 you you are right I made a typo and left out the 2 in front of the x

- anonymous

So I evaluate what I wrote above at zero
but I dont get pi/2, which is the right asnwer

- 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

that should be pretty easy to find ?

- freckles

because we know both tan(0)=0 and 0^2=0

- freckles

so ?=0

- freckles

\[(f^{-1})'(3)=\frac{1}{f'(0)}\]

- freckles

last thing for you to do is just to plug in 0 into your f' you found

- anonymous

I did multiple times
but I dont get the right answer

- freckles

\[f'(0)=2(0)+\frac{\pi}{2}[\sec(\frac{\pi}{2}(0))]^2\]

- freckles

well first the inside of that sec( ) thing you can be simplified
pi/2*0=0
so sec(0)=?

- freckles

\[(f^{-1})'(3)=\frac{1}{f'(0)}=\frac{1}{2(0)+\frac{\pi}{2}[\sec(0)]^2}\]

- freckles

you know cos(0)=1 so sec(0)=?

- freckles

hint sec and cos are reciprocal functions of each other

- anonymous

reciprocal is synonym of inverse?

- freckles

it depends what inverse you are talking about
multiplicative inverse yeah

- freckles

like sec(x)=1/cos(x)
so sec(0)=1/cos(0)

- freckles

1/cos(x) is reciprocal of cos(x)

- freckles

some people just like to call it the flipping of

- anonymous

but it is sec(0)^2
not sec(0)

- anonymous

I meant
sec^2(0)

- 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

1/(a/b) means just flip the number in the denominator

- freckles

so 1/(a/b) is b/a

- freckles

are you cool?

- freckles

still processing?

- anonymous

Thanks man, or girl?

- anonymous

I just made an algebra error

- anonymous

I got it now, but this problem is a pain in the behind

- freckles

I might be a she.
And also it gets way more fun with more practice.

- anonymous

I think my algebra is rusty

- freckles

it seems like everyone is working out their algebra kinks in calculus

- freckles

so no worries

- anonymous

tell me about it

- anonymous

Any one out there who can give me a hand doing integrals

- freckles

post a new question so people can see it :)

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