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

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Answer 1 · 2017-05-08T12:49:15

#f'(x) = lnx(lnx+2)#
#f''(x) = 2/x(lnx+1)#

#f(x) = xln^2x#

#f'(x) = x*2lnx*1/x + ln^2x*1# [Product rule, chain rule and power rule]

#= 2lnx+ln^2x#

#= lnx(lnx+2)#

#f''(x) = lnx*d/dx(lnx+2) +d/dx lnx* (lnx+2)# [Product rule]

#= lnx(1/x+0) + 1/x*(lnx+2)# [Standard differential]

#= 1/x*(2lnx+2)#

#= 2/x(lnx+1)#

Answer 2 · 2017-05-08T13:00:18

Use the product rule to get: #y'=2ln(x)+ln(x)^2#
and the same for getting #y''=2/x(ln(x)+1)#

Step 1. Break the original into two functions

#u=x# and #v=ln(x)^2#

Step 2. Differentiate each #u# and #v#

#d/dx(u)=d/dx(x)=1#

#d/dx(v)=d/dx(ln(x)^2)=2*ln(x)*1/x#

Step 3. Use the Product Rule to solve

Product Rule: #dy/dx=u (dv)/dx+v (du)/dx#

Plugging in: #dy/dx=x*2ln(x)1/x+ln(x)^2*1#
#dy/dx=2ln(x)+ln(x)^2=ln(x)(2+ln(x))#

Step 4. Differentiate the answer from Step 3.
#(d^2y)/dx^2=ln(x)1/x+(2+ln(x))1/x#
#=ln(x)/x+2/x+ln(x)/x#
#=1/x(2ln(x)+2)#
#=2/x(ln(x)+1)#