IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: zabady on April 15, 2010, 01:34:42 am
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i have this doubt in the proof of a^x=INa
i kno that y=a^x
log both sides =
INy=xINa
differentiation of iny is easy but when it comes to differentian of xINa i do it by product rule which is X(1/a)+INa
in the book the ignored the x(1/a) thats where i dnt understand i attached the explanation of the book
my second doubt is attached too
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Xlna is not a product. Lna is a constant
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aha so a constant we would take it as a zero when we differentiate it i have still 1 doubt which is the second picture
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FOR the first doubt , u dont have to know how to derive it
just memories it like a^x = dy/dx=a^xlna
and for the second doubt, u differentiate the whole equ.
2x+8y(dy/dx)-6-16(dy/dx)=0
make dy/dx subject of the formula
dy/dx=(6-2x)/(8y-16)
dy/dx=0
so , x=3
now substitute the x value in the original equ and solve
y= 1 or 3
so the coordinates are (3,1) , (3,3)
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ooooh dam i kno that one i gave the wrong question look at the one i attached here i tried solving it same way as the one u just solved but it gives t=1 and in question they say t is not equal to one so how to solve it ???
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ooooh dam i kno that one i gave the wrong question look at the one i attached here i tried solving it same way as the one u just solved but it gives t=1 and in question they say t is not equal to one so how to solve it ???
okay
dx/dt= 1/(1-t)^2 & dy/dt= t(2-t)/(1-t)^2
so divide dy/dt by dx/dt to give u dy/dx
so , dy/dx= t(2-t)
t= 0 or 2
so the coordinates are
(0,0) , (-2,-4)
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thaaankz so much man sry for wastin ur time it waz easy thankz again mannn ;D ;D ;D ;D ;D ;D ;D ;D ;D
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no problem ;)
im here to help :)
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thanks T.Q
+rep
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thanx ~ A.F ~