How do you find the first and second derivative of #(lnx)^2#?

1 Answer
Jan 12, 2018

Answer:

#"see explanation"#

Explanation:

#"differentiate using the "color(blue)"chain rule"#

#"given "y=f(g(x))" then"#

#dy/dx=f'(g(x)xxg'(x)larrcolor(blue)"chain rule"#

#rArrd/dx((lnx)^2)#

#=2lnx xxd/dx(lnx)=(2lnx)/xlarrcolor(blue)"first derivative"#

#"differentiate first derivative to obtain second derivative"#

#d/dx((2lnx)/x)#

#"differentiate using the "color(blue)"product rule"#

#"given "y=g(x)h(x)" then"#

#dy/dx=g(x)h'(x)+h(x)g'(x)larrcolor(blue)"product rule"#

#"express "(2lnx)/x=2lnx .x^-1#

#g(x)=2lnxrArrg'(x)=2/x#

#h(x)=x^-1rArrh'(x)=-x^-2=-1/x^2#

#rArrd/dx(2lnx .x^-1)#

#=-2lnx . 1/x^2+1/x . 2/x#

#=2/x^2-(2lnx)/x^2larrcolor(blue)"second derivative"#