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प्रश्न
Differentiate the following w.r.t.x:
`(x^2 + 2)^4/(sqrt(x^2 + 5)`
उत्तर
Let y = `(x^2 + 2)^4/(sqrt(x^2 + 5)`
Differentiating w.r.t.x, we get
`"dy"/"dx" = "d"/"dx"[(x^2 + 2)^4/(sqrt(x^2 + 5))]`
`"dy"/"dx" = (sqrt(x^2 + 5)."d"/"dx"(x^2 + 2)^4 - (x^2 + 2)^4."d"/"dx"(sqrt(x^2 + 5)))/(sqrt(x^2 + 5))^2`
`"dy"/"dx" = (sqrt(x^2 + 5) × 4(x^2 + 2)^3."d"/"dx"(x^2 + 2) - (x^2 + 2)^4 × 1/(2(sqrt(x^2 + 5)))."d"/"dx"(x^2 + 5))/(x^2 + 5)`
`"dy"/"dx" = (sqrt(x^2 + 5) × 4(x^2 + 2)^3.(2x + 0) - (x^2 + 2)^4/(2sqrt(x^2 + 5)) × (2x + 0))/(x^2 + 5)`
`"dy"/"dx" = (8x(x^2 + 5)(x^2 + 2)^3 - x(x^2 + 2)^4)/(x^2 + 5)^(3/2)`
`"dy"/"dx" = (x(x^2 + 2)^3[8(x^2 + 5) - (x^2 + 2)])/(x^2 + 5)^(3/2)`
`"dy"/"dx" = (x(x^2 + 2)^3(8x^2 + 40 - x^2 - 2))/(x^2 + 5)^(3/2)`
`"dy"/"dx" = (x(x^2 + 2)^3(7x^2 + 38))/(x^2 + 5)^(3/2)`.
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