What is the derivative of #ln x^5#?

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Answer 1 · 2016-10-17T20:21:09

#5/x#

#d/dx(lnx^5) = d/dx(5lnx) = 5d/dx(lnx) = 5*1/x = 5/x#

Answer 2 · 2016-10-17T21:30:02

#5/x#

If you forget the logarithm rule #log_a(b^c)=clog_a(b)#, we can show the same derivative a different way.

You should know that #d/dxln(x)=1/x#.

According to the chain rule, this also tells us that #d/dxln(u)=1/u*(du)/dx#.

Thus, we see that:

#d/dxln(x^5)=1/x^5*d/dxx^5=1/x^5*5x^4=5/x#