Question

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

Answer 1

Any one can help me with this derivative of an inverse

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

why do you say derivative of inverse ?

Answer 6

Do you mean just find derivative?

Answer 7

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.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

I'm still confused why you said inverse

Answer 12

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

Answer 13

yes freckles

Answer 14

what is the a

Answer 15

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

Answer 16

a = 3

Answer 17

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

Answer 18

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

Answer 19

\[(f^{-1})'(3)=\frac{1}{f'(f^{-1}(3))} \\ f^{-1}(3)=? \implies f(?)=3 \\ \text{ so we need \to find ? such that }3=3+?^2+\tan(\frac{\pi ?}{2}) \\ \text{ and yes } f'(x)=0+2x+\frac{\pi}{2} \sec^2(\frac{\pi}{2}x)\]

Answer 20

that should be pretty easy to find ?

Answer 21

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

Answer 22

so ?=0

Answer 23

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

Answer 24

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

Answer 25

I did multiple times
but I dont get the right answer

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

hint sec and cos are reciprocal functions of each other

Answer 31

reciprocal is synonym of inverse?

Answer 32

it depends what inverse you are talking about
multiplicative inverse yeah

Answer 33

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

Answer 34

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

Answer 35

some people just like to call it the flipping of

Answer 36

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

Answer 37

I meant
sec^2(0)

Answer 38

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


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

Answer 39

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

Answer 40

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

Answer 41

are you cool?

Answer 42

still processing?

Answer 43

Thanks man, or girl?

Answer 44

I just made an algebra error

Answer 45

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

Answer 46

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

Answer 47

I think my algebra is rusty

Answer 48

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

Answer 49

so no worries

Answer 50

tell me about it

Answer 51

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

Answer 52

post a new question so people can see it :)