Question

If `sec^-1((7x^3 - 5y^3)/(7^3 + 5y^3)) = "m", "show"  (d^2y)/(dx^2)` = 0.

Answer 1

`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.