What is the first differential of #f(x) = 2^xln(2x)#?

Question

318 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-18T03:18:34

#f'(x) = 2^x(ln2ln(2x)+1/x)#

#f(x) = 2^xln(2x)#

#f'(x) = d/dx2^x * ln(2x) + 2^x * d/dx ln(2x)#

Apply standard differentials and chain rule

#f'(x) = ln2*2^x * ln(2x) + 2^x* 1/(2x)*2#

#= ln2*2^x*ln(2x)+2^x*1/x#

#=2^x(ln2ln(2x)+1/x)#