How do you find the derivative of # (ln(x))^x#?

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Answer 1 · 2017-01-25T11:04:11

#dy/dx=(lnx)^x{1/lnx+ln(lnx)}.#

#"Let "y=(lnx)^x.#

Taking Natural Log. of both sides, & using the Rule of Log. Fun, we get,

#lny=xln(lnx)#

Now, diff.ing both sides w.r.t. #x#, we get,

#d/dx(lny)=d/dx{xln(lnx)}#

#=xd/dx{ln(lnx)}+{ln(lnx)}d/dx{x}[because," the Product Rule]"#

Now, using the Chain Rule :

#(1) : d/dx(lny)=d/dy(lny)(d/dx(y))=1/ydy/dx;#

#(2) : d/dx{ln(lnx)}=1/lnxd/dx(lnx)=(1/lnx)(1/x)=1/(xlnx);#

Therefore, taking #(1) and (2)# into account, we finally have,

#(1/y)(dy/dx)=x{1/(xlnx)}+{ln(lnx)}(1)=1/lnx+ln(lnx)#

# dy/dx=y{1/lnx+ln(lnx)},# and since, #y=(lnx)^x,# we have,

# dy/dx=(lnx)^x{1/lnx+ln(lnx)}.#

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