If #y=(x^3)ln(x^3)# then #dy/dx=#?

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Answer 1 · 2018-04-09T18:01:10

#y=(x^3)ln(x^3)#

#dy/dx=(x^3)'ln(x^3) + (x^3)[ln(x^3)]'#

#dy/dx = 3x^2ln(x^3) +cancel x^3*(1/cancelx^3*3x^2)#

#dy/dx = 3x^2ln(x^3) + 3x^2#

#dy/dx = 3x^2(ln(x^3) + 1)#

Answer 2 · 2018-04-09T18:02:57

Use #ln(x^3) = 3lnx# to write:

#y = 3x^3lnx#

Now use the product rule to differentiate.

#y' = 9x^2lnx + 3x^3 * 1/x# And simplify

#y' 9x^2lnx+3x^2 = 3x^2(1+3lnx)#