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Author Topic: Re: IBOct/Nov 2009...How was the exams for you?  (Read 2405 times)

Offline astarmathsandphysics

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Re: IBOct/Nov 2009...How was the exams for you?
« 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 )
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Offline astarmathsandphysics

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Re: CIE Oct/Nov 2009...How was the exams for you?
« Reply #1 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 )
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Offline falafail

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Re: CIE Oct/Nov 2009...How was the exams for you?
« Reply #2 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:
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Offline astarmathsandphysics

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Re: CIE Oct/Nov 2009...How was the exams for you?
« Reply #3 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)
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Offline astarmathsandphysics

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Re: CIE Oct/Nov 2009...How was the exams for you?
« Reply #4 on: November 10, 2009, 04:15:46 pm »
Someone asked how to differentiate the inverse trig functions
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Offline falafail

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Re: CIE Oct/Nov 2009...How was the exams for you?
« Reply #5 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
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Offline Tyserius

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Re: CIE Oct/Nov 2009...How was the exams for you?
« Reply #6 on: November 10, 2009, 04:21:48 pm »
LOL astar xD I think it's another thread xD
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Offline astarmathsandphysics

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Re: CIE Oct/Nov 2009...How was the exams for you?
« Reply #7 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
« Last Edit: November 10, 2009, 04:33:12 pm by astarmathsandphysics »
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