What is the derivative of #x^(1/x)#?

1 Answer
VNVDVI
Apr 2, 2018

#dy/dx=x^(1/x)((1-lnx)/x^2)#

Explanation:

In these situations where a function is raised to the power of a function, we'll use logarithmic differentiation and implicit differentiation as follows:

#y=x^(1/x)#

#lny=ln(x^(1/x))#

From the fact that #ln(a^b)=blna#:

#lny=lnx/x#

Differentiate (the left side will be differentiated implicitly):

#1/y*dy/dx=(1-lnx)/x^2#

Solve for #dy/dx#:

#dy/dx=y((1-lnx)/x^2)#

Recalling that #y=x^(1/x)#:

#dy/dx=x^(1/x)((1-lnx)/x^2)#

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