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IGCSE/GCSE/O & A Level/IB/University Student Forum > Qualification > IB > Subject Doubts > Re: IBOct/Nov 2009...How was the exams for you?
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Author Topic: Re: IBOct/Nov 2009...How was the exams for you?  (Read 258 times)
astarmathsandphysics
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« 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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astarmathsandphysics
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« 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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« Reply #2 on: November 10, 2009, 04:13:25 PM »
Quote from: Tyserius on November 10, 2009, 04:07:02 PM
Haha thanks Cheesy 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 Cheesy

oh wow. awesome Grin


@astar what D:
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astarmathsandphysics
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« 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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« Reply #4 on: November 10, 2009, 04:15:46 PM »
Someone asked how to differentiate the inverse trig functions
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« 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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« Reply #6 on: November 10, 2009, 04:21:48 PM »
LOL astar xD I think it's another thread xD
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« 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 » Logged
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