Let, f(x)=ln(x^8)/x^2=(8lnx)/x^2.....[because, ln a^m=mlna],
:. f(x)=8*x^-2*lnx.
By the Product Rule, then, we get,
f'(x)=8{x^-2d/dx(lnx)+(lnx)d/dx(x^-2)},
=8{x^-2*1/x+(-2*x^(-2-1))lnx},
=8{x^-1-2x^-3*lnx},
=8{1/x-(2lnx)/x^3},
rArr f'(x)=8/x^3(x^2-2lnx).
Knowing that, f''(x)={f'(x)}', we have,
f''(x)={8(x^-1-2x^-3*lnx)}',
=8(x^-1-2x^-3*lnx)',
=8(x^-1)'-8*2(x^-3*lnx)',
=8(-1*x^(-1-1))-16{x^-3d/dx(lnx)+(lnx)d/dx(x^-3)},
=-8x^-2-16{x^-3*1/x+(-3*x^(-3-1))lnx},
=-8/x^2-16(1/x^2-(3lnx)/x^4),
=-8/x^2-16/x^2+(48lnx)/x^4,
=-24/x^2+(48lnx)/x^4.
rArr f''(x)=24/x^4(2lnx-x^2).
Enjoy Maths.!