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

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Answer 1 · 2017-09-25T11:56:06

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

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

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.!