How do you differentiate #f(x)=lnx*e^(x-1)# using the product rule?

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Answer 1 · 2015-12-15T02:08:52

#f'(x)=(e^(x-1)(1+xlnx))/x#

According to the product rule:

#f'(x)=e^(x-1)d/dx[lnx]+lnxd/dx[e^(x-1)]#

Find each derivative separately.

#d/dx[lnx]=1/x#

This requires the chain rule:

#d/dx[e^(x-1)]=e^(x-1)d/dx[x-1]=e^(x-1)*1=e^(x-1)#

Plug these back in.

#f'(x)=e^(x-1)(1/x)+lnx(e^(x+1))#

#f'(x)=(e^(x-1))/x+(xlnx(e^(x-1)))/x#

#f'(x)=(e^(x-1)(1+xlnx))/x#