How do you find the derivative of #y=xlnx#?

You'll need the product rule for this one. The product rule is given by:

#=(f(x)*g(x))'=f'(x)g(x)+f(x)g'(x)#

Nel caso di #y=xln(x)#, #f(x)=x# e #g(x)=ln(x)#.

First we take the derivative of #f(x)#. The derivative of a single variable (no coefficient, not raised to any power) is #1#. We leave #g(x)# alone, so the first half of the derivative is simply #1*ln(x)=ln(x)#.

Then we take the derivative of #g(x)#. The derivative of #ln(x)# is #1/x#. We leave #f(x)# alone, so the second have of the derivative is #x*1/x=1#.

Mettendo tutto insieme, otteniamo #y'=ln(x)+1#.

Spero che questo ti aiuti!