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प्रश्न
If `sec^-1((7x^3 - 5y^3)/(7^3 + 5y^3)) = "m", "show" (d^2y)/(dx^2)` = 0.
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उत्तर
`sec^-1((7x^3 - 5y^3)/(7^3 + 5y^3))` = m
∴ `(7x^3 - 5y^3)/(7x^3 + 5y^3)` = sec m =k ...(Say)
∴ 7x3 – 5y3 = 7kx3 + 5ky3
∴ (5k + 5)y3 = (7 – 7k)x2
∴ `y^3/x^3 = (7 - 7k)/(5k + 5)`
∴ `y/x = ((7 - 7k)/(5k + 5))^(1/3)` = p, where p is a constant
∴ `"d"/"dx"(y/x) = "d"/"dx"(p)`
∴ `(x"dy"/"dx" - y "d"/"dx"(x))/(x^2)` = 0
∴ `x"dy"/"dx" - y xx 1` = 0
∴ `x"dy"/"dx"` = y
∴ `"dy"/"dx" = y/x` ...(1)
∴ `(d^2y)/(dx^2) = "d"/"dx"(y/x)`
= `(x"dy"/"dx" - y "d"/"dx"(x))/(x^2)`
= `(x(y/x) - y xx 1)/(x^2)` ...[By (1)]
= `(y - y)/x^2`
= `0/x^2`
= 0
Note : `"dy"/"dx" = y/x. "where" y/x` = p.
∴ `"dy"/"dx"` = p, where p is a constant.
∴ `(d^2y)/(dx^2) = "d"/"dx"(p)` = 0.
