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Question #732ab

1 Answer

sankarankalyanam

Oct 23, 2017

$y’ = 4/sqrt(x^2 - 2)$

Explanation:

d /(dx) (cos^-1 x) = -1/sqrt(1-x^2)#

$y = cos ^-1 (2/x^2)$
$:. y’ = (-1/(sqrt(1-(2/x^2))) * (-4/x))$

$y’ = (4/cancel(x))(cancel(x )/ (sqrt(x^2 - 2)))$

$y’ = 4 / sqrt(x^2 - 2)$

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