How do you find the first and second derivative of ln (x^8)/ x^2ln(x8)x2?

1 Answer
Sep 25, 2017

(1) :" The First Derivative is, "8/x^3(x^2-2lnx).(1): The First Derivative is, 8x3(x22lnx).

(2) :" The Second Derivative is, "24/x^4(2lnx-x^2).(2): The second Derivative is, 24x4(2lnxx2).

Explanation:

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.!