Question

sin(inv) {(a+bcosx)/(b+acosx)}

Answer 1

Differentiate...

Answer 2

i was scared that you would ask to calculate the trigonometric value.

Answer 3

is there a possible sub..

Answer 4

sub??

you know how to differentiate sin-1x right??

Answer 5

yes i know the long method but isnt there a shorter sub method..

Answer 6

http://www.themathpage.com/acalc/inverse-trig.htm#arcsin

and you know how to use chain rule and quotient rule right??

Answer 7

of course i,know all those and i have also found the answer using the quotient rule ,but my main q was that is there a possible substitution to reduce the time reqd to do the sum...

Answer 8

\[\sin^{-1}\left(\frac {a+b\cos(x)}{b+a\cos(x)}\right)\]

Answer 9

yes@satellite thats the one..

Answer 10

i was just checking to see if the inside piece was something snappy, but it doesn't look like it, so you just have to grind it til you find it

Answer 11

to make it a little quicker you could, is suppose, find
\[\frac{d}{dx}\frac{a+bx}{b+ax}=\frac{b^2-a^2}{(a+bx)^2}\] and then use the chain rule to find the derivative of the inside part easier

Answer 12

whole thing looks like a colossal pain

Answer 13

yeah it does seem so,think how am i supposed to do these sums under five minutes without any error in the exams
okay,thanks @satellite,@experimentX

Answer 14

yes, doesn't look like there is a short cut. i wolframed it and it is just what you think it is
take the derivative of the inside piece, get
\[\frac{(b^2-a^2)\cos(x)}{(a+b\cos(x))^2}\] then take derivative of arcsine, and use chain rule
looks like what you expect
http://www.wolframalpha.com/input/?i=arcsine%28%28a%2Bbcos%28x%29%29%2F%28b%2Ba*cos%28x%29%29%29

Answer 15

looks ugly ... as expected.