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Question
`"d"/("d"x) [sin(1 - x^2)]^2` = ______.
Options
`-2x(1 - x^2)cos(1 - x^2)`
`-4x(1 - x^2)cos(1 - x^2)`
`4x(1 - x^2)cos(1 - x^2)^2`
`-2(1 - x^2)cos(1 - x^2)^2`
MCQ
Fill in the Blanks
Solution
`"d"/("d"x) [sin(1 - x^2)]^2` = `-4x(1 - x^2)cos(1 - x^2)`.
Explanation:
`"d"/("d"x) [sin(1 - x^2)]^2`
= `cos(1 - x^2)^2 * "d"/("d"x)(1 - x^2)^2`
= `cos(1 - x^2)^2 * 2(1 - x^2) * "d"/("d"x)(1 - x^2)`
= `cos(1 - x^2)^2 * 2(1 - x^2) * (-2x)`
= `-4x(1 - x^2) cos(1 - x^2)^2`
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