IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

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The Derivative of sec^ -1 [1/(2x^2-1)] with respect to ^{[ ]}

1-x^2 at x=0.5

solve this!!!

ganesha1991

is the question 1/sec [ 1/ [2x² -1]

vaibhav_hyderabad1991

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@ ganesha1991
it is secant inverse
i am sorry if i confused you

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@ vaibhav_hyderabad1991
can you pls explain answer is correct!

deedee

put x=cos @

thn then

differentiate

u'll get

y=sec^-1(sec2@)

thus

y=2@

thus dy/d@=2

then wrt 2 (1-x^@)^1/2

wich u get as sin@

dx/d@=cos @

thus dy/dx=2/cos@

cos@=x=.5

thsu substituettin u egt d ansewr

ganesha1991

y=(sec^-1 [ 1/ (2x^2 - 1) ]
x=cosx
y = sec^-1 [ 1/ 2cos^2-1 ]
2cos^2 x - 1 = cos2x
therefore
y = sec^-1 [ 1/cos2x ]
y= sec^-1 [ sec2x]
therefore
y = 2x
dy/dx = 2 ------------------------(1)
y= root1-x^2
dy'/dx =[ 1/ {2root(1-x^2)} ] * (-2x)
= [-x/(root(1-x^2) ] -----------------------------(2)
(1) / (2)
dy/dy' = 2 / { [-x/(root(1-x^2) ) ]
x=1/2
dy/dy' = 2/ [ -1/2[ root (3/4)]
= -2root3

therefore the answer is -2

3

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but answer is 4

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yes deede i got it
thanks!!!

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