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Question
If y = `1/sqrt(3x^2 - 2x - 1)`, then `("d"y)/("d"x)` = ?
Options
`(-2)/3 (3x - 2) (3x^2 - 2x - 1)^((-3)/2)`
`(-3)/2 (3x - 2) (3x^2 - 2x - 1)^((-3)/2)`
`(3x - 1) (3x^2 - 2x - 1)^((-3)/2)`
`-(3x - 1) (3x^2 - 2x - 1)^((-3)/2)`
Solution
`bb(-(3x - 1) (3x^2 - 2x - 1)^((-3)/2))`
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