How do you differentiate #f'(x) = 2x log(x) + x#?

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Answer 1 · 2015-10-14T01:40:32

#(d^2y)/dx^2 = 2/(ln(10)) + 2log(x) + 1#

We have, assuming that #log(x)# is the base 10 logarithm,

#dy/dx = 2xlog(x) + x#

To find #(d^2y)/dx^2# we need to use the product rule

#(d^2y)/dx^2 = 2xd/dx(log(x)) + log(x)d/dx(2x) + d/dx(x)#

#(d^2y)/dx^2 = 2xd/dx(log(x)) + 2log(x) + 1#

We can rewrite #log(x)# as #ln(x)/ln(10)#

#(d^2y)/dx^2 = (2x)/ln(10)d/dx(ln(x)) + 2log(x) + 1#

#(d^2y)/dx^2 = (2x)/(xln(10)) + 2log(x) + 1#

#(d^2y)/dx^2 = 2/(ln(10)) + 2log(x) + 1#

If you meant the natural log, just replace #ln(10)# for #ln(e)# or #1#