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प्रश्न
Find `("d"y)/("d"x)`, if y = `root(3)(((3x - 1))/((2x + 3)(5 - x)^2)`
उत्तर
y = `root(3)(((3x - 1))/((2x + 3)(5 - x)^2)`
= `(3x - 1)^(1/3)/((2x + 3)^(1/3)*(5 - x)^(2/3)`
Taking logarithm of both sides, we get
log y = `log[(3x - 1)^(1/3)/((2x - 3)^(1/3)*(5 - x)^(2/3))]`
= `log(3x - 1)^(1/3) - [log(2x + 3)^(1/3) + log(5 - x)^(2/3)]`
= `1/3log(3x - 1) - [1/3 log(2x + 3) + 2/3log(5 - x)]`
Differentiating both sides w.r.t. x, we get
`"d"/("d"x)(log y) = 1/3*"d"/("d"x)[log(3x - 1)] - 1/3*"d"/("d"x)[log(2x + 3)] - 2/3*"d"/("d"x)[log(5 - x)]`
∴ `1/y*("d"y)/("d"x) = 1/3*1/((3x - 1))*"d"/("d"x)(3x - 1) - 1/3*1/((2x + 3))*"d"/("d"x)(2x + 3) - 2/3*1/((5 - x))*"d"/("d"x)(5 - x)`
∴ `1/y*("d"y)/("d"x) = 1/(3(3x - 1)) xx (3 - 0) - 1/(3(2x + 3)) xx (2 + 0) - 2/(3(5 - x)) xx (0 - 1)`
∴ `1/y*("d"y)/("d"x) = 1/(3x - 1)- 2/(3(2x + 3)) + 2/(3(5 - x))`
∴ `("d"y)/("d"x) = y/3[3/(3x - 1) - 2/(2x + 3) + 2/(5 - x)]`
∴ `("d"y)/("d"x) = 1/3 root(3)((3x - 1)/((2x + 3)*(5 - x)^2)) [3/(3x - 1) - 2/(2x + 3) + 2/(5 - x)]`
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