Question

Differentiate the inverse trig function.

Answer 1

\[\sqrt{x^2-1}arcsecx\]

Answer 2

product rule (uv), aware of that ?

Answer 3

yes, but the sqrt(x^2-1), does that need to be chain ruled somehow?

Answer 4

yup, chain rule is required there

Answer 5

\[\sqrt{x^2-1}(\frac{ 1 }{x \sqrt{x^2-1} })+\sec^{-1}(\frac{ 1 }{ 2 }(x^2-1))^{-1/2}\]

Answer 6

that cant be right

Answer 7

so, if i chain rule

Answer 8

what did u do in 2nd term ?

Answer 9

1st term is correct, and u can cancel something there, what ?

Answer 10

\[\frac{ 1 }{ 2 }(x^2-1)^{-1/2}(\sqrt{x^2-1})(\frac{ 1 }{ x \sqrt{x^2-1} }) + arcsecx \frac{ 1 }{ 2 }(x^2-1)^{-1/2}\]

Answer 11

is that the chain rule?

Answer 12

no...

Answer 13

first , did u apply product correctly?

Answer 14

fg'+gf'

Answer 15

f = sqrt(x^2-1) g=arcsecx

Answer 16

g= arcsec x right ?

Answer 17

ok,
so what is fg' = ?

Answer 18

\[\sqrt{x^2-1}(\frac{ 1 }{ x \sqrt{x^2-1} })\]

Answer 19

can u cancel something ?

Answer 20

which is 1/x

Answer 21

yes!
to be precise its 1/|x|
so now you are completely done with fg'
it equals 1/|x|
now what is gf' ?

Answer 22

\[arcsecx(\frac{ 1 }{ 2 }(x^2-1))^{-1/2}(2)\]

Answer 23

chain rule ok?

Answer 24

2 or 2x in the end ?

Answer 25

yes, chain rule applied correctly.
but d/dx(x^2-1) is 2 or 2x ??

Answer 26

oh 2x my bad

Answer 27

:0

Answer 28

now put everything together
fg'+f'g =?

Answer 29

remember to cancel 2 from numerator and denominator of 2nd term

Answer 30

\[\frac{ 1 }{ |x| }+\frac{ 2xarcsecx }{ 2\sqrt{x^2-1} }\]

Answer 31

*2 can be cancelled*

Answer 32

oh ya duh :)

Answer 33

so, did u get your required final answer ?

Answer 34

\[\frac{\sqrt{x^2-1}+xarcsecx }{ x \sqrt{x^2-1} }\]

Answer 35

wouldn't it be x^2 in numerator ? before arc....
you could have left it in 2 terms form also

Answer 36

oh ya, so its x^2arcsecx

Answer 37

forgot to cross multiply the x

Answer 38

is that it though?

Answer 39

yup, thats it.

Answer 40

!!!!!

Answer 41

@hartnn you saved the day yet again!!! :)

Answer 42

as usual...welcome ^_^

Answer 43

\[\frac{ \sqrt{x^2-1}+x^2arcsecx }{ x \sqrt{x^2-1} }\]

Answer 44

yup.

Answer 45

weeew! thanks so much!!!