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

Question

If sec^-1\(\left(\frac{7x^3 - 5y^3}{7x^3 + 5y^3}\right) = m,\) show that d²y/dx² = 0

sec^-1((7x³ - 5y³)/(7x³ + 5y³)) = m,

Answer 1

sec^-1((7x³ - 5y³)/(7x³ + 5y³)) = m

 sec^-1\(\left(\frac{7x^3 - 5y^3}{7x^3 + 5y^3}\right) = m\)

\(\therefore\) dy/dx = p, where p is a constant.