what is the derivative of tan(1)?

- anonymous

what is the derivative of tan(1)?

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- schrodinger

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- yuki

0

- yuki

because tan(1) is a constant

- anonymous

okay then i have to post the actual question bc i am having trouble

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## More answers

- yuki

okay, shoot it

- anonymous

use this formula |E_{n}(x)|\le\frac{M}{(n+1)!}|x-a|^{{n+1}} find the bound for the error in approximating the quantity with a third degree polynomial for tan 1, f(x) =tanx about x=0

- yuki

I have no idea what you plugged in lol
are you asking us to find what tan(1) is using the McLaurin series with degree 3 ?

- anonymous

i have to find the error bound using the lagrange error bound for P(x)

- anonymous

that the formula that i typed

- yuki

okay, then let's do it step by step.
did you find the Taylor (in our case McLaurin because the center is at x=0) Series of tan (x) ?

- anonymous

no not yet

- yuki

to use the Lagrange error bound you need the (n+1)th term in the Taylor polynomial for tan(x). since we are approximating tan(x) by a 3rd degree, we need the 4th term of the Taylor polynomial.

- anonymous

okay

- yuki

are you comfortable finding it ?

- anonymous

i found it i got -6( 1+ x^2)^-4

- anonymous

oh i did it wrong i wrote down the wrong derivative

- anonymous

i need helpp

- yuki

the derivative of tan(x) = sec^2(x)
which makes it really hard to find te 4th term.
take your time

- anonymous

i did find it i got -24 sinx/(cosx)^5

- yuki

now we need to find the upper bound of that term.
the problem here is that -24sin(x)/(cosx)^5 has no bound
because when cos(x) is close to 0, this number becomes arbitrarily big.

- anonymous

so what do i do??

- yuki

give me one second

- anonymous

ok

- yuki

were you given any interval ?

- anonymous

it just says x=0 and tan (1) so 0

- yuki

nath lol
that info was actually very important lmao

- yuki

since x is between 0 and 1, we just need to find the upper bound of -24 sin/cos^5 between those intervals

- yuki

cosine decreases as x increases from 0 to 1 (radians)
and sine increases as x increases from 0 to 1
so the larges number M can be is found by plugging in x =1
into the fourth derivative of tan(x)
just to let you know I am not really sure if it is -24sin(x)/cos^5(x), but if you are 100% sure we can proceed

- anonymous

is there anyway you could do it really quickly just to make sure its correct cause we are make an effort to do the problem

- yuki

I will try to find it, then.
try to trust me for the derivative.
even if we don't get the right answer the concept is solid.
give me a minute to find f(^4)(x)

- anonymous

k ty

- yuki

\[f = tanx\]
\[f' = \sec^2x\]
\[f" = 2\sec^2x tanx\]
\[f^{(3)} = 4secx tanx + 2 \sec^4x\]
\[f^{(4)} = 4secx \tan^2x + 4 \sec^3x + 8\sec^4tanx\]

- yuki

let me see if I can simplify it

- anonymous

kk

- yuki

I think this one looks best among others
\[f^{(4)} = {2\sin^2(2x) +4\cos^2(x) +8\sin(x) \over \cos^5(x)}\]

- yuki

so M would be
2sin^2(2)+4cos^2(1) + 8sin(1) over cos(1)

- anonymous

the thing is i said the derivative of tan x is 1/cos^2 x and i got something completely different from you

- anonymous

okay lets move on so how do i know what to plug into x

- yuki

let's try that as well
\[f' = { 1 \over \cos^2(x)}\]
\[f" = {2\sin(x) \over \cos^4(x)}\]
which simplyfies to 2sec^2(x)tan(x)
\[f^{(3)} = {2\cos^4(x) +6\cos^2(x)\sin^2(x) \over \cos^6(x)}\]
I don't think I really want to do the next one ...

- yuki

basically your idea is
"what is the maximum value of f^(n+1) in the given interval ?"
if you are using a Taylor series with center a to apprx, tan(2),
the interval is from a < x < 2
if they want you to evaluate tan(x) from -3 < x < 5 for example,
you will pick the x that maximizes f^(n+1) between -3 and 5

- anonymous

so for this problem what my max varies for cos and and sin

- anonymous

bc their max is different

- yuki

so if we used the f^(4) I got, what you can say is
I cannot calculate sin(1) or cos(1) , but I know that
1 radian is less than 60 degrees and greater than 0.
to maximize the expression that I have,
I need to make the numerator as small as possible and
the denominator as large as possible.
so since 0

- yuki

ops nvm

- yuki

i couldn't get a nice number ...

- yuki

anyway, I don't feel like I was of much help,
but the Idea is this.
once you get a valid M,
you start solving for n.
since it will involve factorials, all you have to do is to plug in
a couple of integers to see when it gives you a true statement.
since tangent is not an easy function to calculate the derivatives, I am not 100% sure with my calculations either.

- anonymous

alright ty u very much

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