質問: How do you find the first and second derivative of #ln (x^8)/ x^2#? ===================================================================
回答: # (1) :" The First Derivative is, "8/x^3(x^2-2lnx).# # (2) :" The Second Derivative is, "24/x^4(2lnx-x^2).# #### 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.!**