IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => IB => Commerce => Topic started by: astarmathsandphysics on November 10, 2009, 04:09:08 pm

Title: Re: IBOct/Nov 2009...How was the exams for you?
Post by: astarmathsandphysics on November 10, 2009, 04:09:08 pm: $y=cos^{-1} x$
$cosy=x$
differentiate with respect to y to get
$-siny=\frac {dx}{dy}$
$\frac {dy}{dx}=-1/siny=-1/sqrt(1-cos^2 y)=-1/sqrt(1-x^2 )$
Title: Re: CIE Oct/Nov 2009...How was the exams for you?
Post by: astarmathsandphysics on November 10, 2009, 04:12:13 pm: $y=sin^{-1} x$
$siny=x$
differentiate with respect to y to get
$cosy=\frac {dx}{dy}$
$\frac {dy}{dx}=1/cosy=1/sqrt(1-sin^2 y)=1/sqrt(1-x^2 )$
Title: Re: CIE Oct/Nov 2009...How was the exams for you?
Post by: falafail on November 10, 2009, 04:13:25 pm: Quote from: Tyserius on November 10, 2009, 04:07:02 pm
Haha thanks :D Trying out for Harvard, Yale, Princeton, MIT, Boston U and Wharton D: Gonna do a double degree, Math and Economics. Then I might proceed for Masters in Actuarial Science :D

oh wow. awesome ;D

@astar what D:
Title: Re: CIE Oct/Nov 2009...How was the exams for you?
Post by: astarmathsandphysics on November 10, 2009, 04:14:55 pm: $y=tan^{-1} x$
$tany=x$
differentiate with respect to y to get
$sec^2 y=\frac {dx}{dy}$
$\frac {dy}{dx}=1/sec^2 y=1/(1+tan^2 y) =1/(1+x^2)$
Title: Re: CIE Oct/Nov 2009...How was the exams for you?
Post by: astarmathsandphysics on November 10, 2009, 04:15:46 pm: Someone asked how to differentiate the inverse trig functions
Title: Re: CIE Oct/Nov 2009...How was the exams for you?
Post by: falafail on November 10, 2009, 04:16:44 pm: Quote from: astarmathsandphysics on November 10, 2009, 04:15:46 pm
Someone asked how to differentiate the inverse trig functions

oh, okay. i thought maybe you were posting in the wrong thread XD
Title: Re: CIE Oct/Nov 2009...How was the exams for you?
Post by: Tyserius on November 10, 2009, 04:21:48 pm: LOL astar xD I think it's another thread xD
Title: Re: CIE Oct/Nov 2009...How was the exams for you?
Post by: astarmathsandphysics on November 10, 2009, 04:24:37 pm: $y=sec^{-1} x$
$secy=x$
differentiate with respect to y to get
$tany secy=\frac {dx}{dy}$
$\frac {dy}{dx}=1/tanysecy=1/sqrt(sec^2 y -1) secy =1/sqrt(x^2 -1)x$

$y=cosec^{-1} x$
$cosecy=x$
differentiate with respect to y to get
$-cotycosecy=\frac {dx}{dy}$
$\frac {dy}{dx}=-1/cotycosecy=-1/sqrt(cosec^2 y -1) cosecy =-1/sqrt(x^2 -1)x$