Question

derivative of arcsin((5x+12(sqrt(1-x^2)))/13)

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

\[\frac{d}{dx} \sin^{-1} u = \frac{1}{\sqrt{1-u^2}} \frac{du}{dx}\]

Answer 3

lae x=sin theta
5x+12sqrt(1-x^2)/13=sin(theta+arctan (12/5)
so ans arctan 12/5

Answer 4

its jsut a really messy chain rule

Answer 5

lae x=sin theta 5x+12sqrt(1-x^2)/13=sin(theta+arctan (12/5)
so ans is derivative of theta+ arctan 12/5= sin^-1x+c
1/sqrt(1-x^2)

Answer 6

see if u sub. this x=sin theta then it is reduced greatly...

Answer 7

^ I dont have a cule what u did at all lols

Answer 8

clue*

Answer 9

I am very sure you have something wrong

Answer 10

\[\sin^{-1} (\frac{5x+12\sqrt{1-x^2}}{13})\]let \[x=\sin \theta\]\[\sqrt{1-x^2}\sqrt{(1-\sin^2 \theta)}=\cos \theta\]\[\frac{5x+12\sqrt{1-x^2}}{13}=\frac{(5\sin \theta+12 \cos \theta)}{13}=\sin(\theta +\tan^{-1} (\frac{12}5)\]\[\sin^{-1}(\sin (\theta+\tan^{-1}(\frac{12}5)))=\theta +\tan^{-1}(\frac{12}5)\]
\[\frac d{dx}(\sin^{-1}x+\tan^{-1}(12/5))=\frac{1}{\sqrt{1-x^2}}\]

Answer 11

lol dont worry , your working still makes no sense , and you kind off got sided tracked and didnt answer the actual at all

Answer 12

for example , the third last line of above post

Answer 13

ohh I think i get third line

Answer 14

but still you are doing too many steps in one line lols

Answer 15

mr elecengineer if havin so much prob frm his workin y dnt u get ans fr me

Answer 16

\[\frac{5\sin \theta +12 \cos \theta }{13}\]let \[\frac5{13}=\cos \alpha , \]so\[\frac{12}{13}=\sin \alpha\]\[\frac5{13}\sin \theta + \frac{12}{13}\cos \theta =\sin \theta \cos \alpha +\cos \theta \sin \alpha=\sin (\theta +\alpha)\]now \[\alpha=\tan^{-1}\frac{12}{13}\]

Answer 17

elecengineer..............don't be so fast....!!!!!!!! it causes accident.!!!!!!!!!!!!

Answer 18

I dont have any specific problem with it , just that I checked wolframa and the answer they give their is massive (obviously the result of chain rules )

Answer 19

http://www.wolframalpha.com/input/?i=arcsin%28%285x%2B12%28sqrt%281-x%5E2%29%29%29%2F13%29

Answer 20

there is no short cut, when you change the variables by substitution then you must use even more chain rules when differentiating

Answer 21

then it becomes so many chain rules that people start making mistakes

Answer 22

look elec...i substituted only to reduce the given problem to a simplified form....

Answer 23

has vaishali understood? thats the big query..

Answer 24

its the fault of that person who makes mistake using chain rule ...not me...

Answer 25

but I dont even get the whole point of your last line\[\frac{d}{dx} (\sin^{-1}(x) + \tan^{-1}( \frac{12}{5}) ) = \frac{1}{\sqrt{1-x^2}} \]

Answer 26

hey guyz stop it

Answer 27

like, isnt that obvious?, and it didnt answer the question

Answer 28

I think you forgot what the actual question was lol

Answer 29

elecengr is right..its actually simplified...theres no needo use chain rule

Answer 30

if u sub x =sint
it simply reduces to arcsin(sin(t + alpha))
which is t + alpha
or
arcsinx + alpha
and its derivative is known to all:)

Answer 31

no i havn't......i solve problems in my own style..not like just coping from wolframalpha or in a conventional way....thats my salient features...

Answer 32

got it vaishali?

Answer 33

we guys dont copy frm websites here...chill..calm down ppl

Answer 34

I dont copy it from wolframa , but when someone does put up a really small answer to something that involves a ton of chain rules then I am a little suspect and I use wolframa to verify the correctness of the answer when I dont feel like doing all the mindless differentiation

Answer 35

and wolframa does not agree with you at all

Answer 36

For what it is worth, a Mathematica 8 solution with comments is attached.