Differentiate the function? f(x) = log13(xe^x)

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Answer 1 · 2017-03-17T15:47:56

#f'(x)=e^xlog13(1+x)#

#f(x)=log13(xe^x)#

Recall that the product rule allows us to differentiate a function that is the product of two functions.

If #h(x)=f(x)*g(x)#
then #h'(x)=f'(x)*g(x)+f(x)*g'(x)#

So, if we have

#f(x)=xe^xlog13#

Using the product rule:

#f'(x)=(1)log13e^x+xe^xlog13#

#f'(x)=e^xlog13(1+x)#

Answer 2 · 2017-03-17T20:15:43

As presented, we have #f(x) = log(13xe^x)#. I'll assume that #log# is natural logarithm.

#d/dx(logu) = 1/u (du)/dx#

So we have #f'(x) = 1/(13xe^x) d/dx(13xe^x)#.

Using the product rule we get

#d/dx(13xe^x) = 13e^x + 13x e^x = 13e^x(1+x)#.

Thus #f'(x) = 1/(13xe^x) * 13e^x(1+x) = (1+x)/x = 1/x+1#.

Answer 3 · 2017-03-17T20:22:12

Perhaps the intended function is #f(x) = log_13(xe^x)# in which case

Use #d/dx(log_b u) = 1/(u lnb) * (du)/dx# to get

#f'(x) = 1/(x e^x ln13) * d/dx(xe^x)#

Now use the product rule to find

#d/dx(xe^x) = e^x+xe^x = e^x(1+x)#.

So we have

#f'(x) = 1/(x e^x ln13) * e^x(1+x) = (1+x)/(xln13)#