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प्रश्न
Differentiate the following w.r.t.x:
`(sqrt(3x - 5) - 1/sqrt(3x - 5))^5`
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उत्तर
Let y = `(sqrt(3x - 5) - 1/sqrt(3x - 5))^5`
Differentiating w.r.t. x,we get,
`"dy"/"dx" = "d"/"dx"(sqrt(3x - 5) - 1/sqrt(3x - 5))^5`
`= 5(sqrt(3x - 5) - 1/sqrt(3x - 5))^4."d"/"dx"(sqrt(3x - 5) - 1/sqrt(3x - 5))`
`= 5(sqrt(3x - 5) - 1/sqrt(3x - 5))^4.["d"/"dx"(3x - 5)^(1/2) - "d"/"dx"(3x - 5)^(-1/2))]`
`= 5(sqrt(3x - 5) - 1/sqrt(3x - 5))^4. [1/2(3x - 5)^(-1/2)."d"/"dx"(3x - 5) - (-1/2)(3x - 5)^(-3/2)."d"/"dx"(3x - 5)]`
`= 5(sqrt(3x - 5) - 1/sqrt(3x - 5))^4. [1/(2sqrt(3x - 5)).(3 × 1 - 0) + 1/(2(3x - 5)^(3/2)).(3 × 1 - 0)]`
`= 5(sqrt(3x - 5) - 1/sqrt(3x - 5))^4. [3/(2(3x - 5)^(1/2)) +3/(2(3x - 5)^(3/2))]`
`= 5(sqrt(3x - 5) - 1/sqrt(3x - 5))^4. 3/2 [1/(3x - 5)^(1/2) + 1/(3x - 5)^(3/2)]`
`= 15/2 (sqrt(3x - 5) - 1/sqrt(3x - 5))^4. [(1 × (3x - 5))/((3x - 5)^(1/2) × (3x - 5)^1) + 1/(3x - 5)^(3/2)]`
`= 15/2 (sqrt(3x - 5) - 1/sqrt(3x - 5))^4. [(1 × (3x - 5))/((3x - 5)^(1/2 + 1)) + 1/(3x - 5)^(3/2)]`
`= 15/2 (sqrt(3x - 5) - 1/sqrt(3x - 5))^4. [(3x - 5)/((3x - 5)^(3/2)) + 1/(3x - 5)^(3/2)]`
`= 15/2 (sqrt(3x - 5) - 1/sqrt(3x - 5))^4. [(3x - 5 + 1)/((3x - 5)^(3/2))]`
`= (15(3x - 4))/(2(3x - 5)^(3/2))(sqrt(3x - 5) - 1/sqrt(3x - 5))^4`.
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