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How do you differentiate #y=ln(3xe^(1-x))#?

1 Answer

Jul 31, 2018

# (1-x)/x#.

Explanation:

#y=ln(3xe^(1-x))#.

Using the usual rules of log function, we have,

#y=ln3+lnx+lne^(1-x), or, #

#y=ln3+lnx+(1-x)lne=ln3+lnx+1-x#.

#:. dy/dx=0+1/x+0-1#.

# rArr dy/dx=(1-x)/x#, as Respected Sonnhard has derived!

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