How do you find the derivative of #x^(1/x)#?

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Answer 1 · 2018-04-01T19:47:33

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

When dealing with a function raised to the power of a function, logarithmic differentiation becomes necessary.

Let #y=x^(1/x)#

Then,

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

Recalling that #ln(x^a)=alnx:#

#lny=1/xlnx#

#lny=lnx/x#

Now, differentiate both sides with respect to #x,# meaning that the left side will be implicitly differentiated:

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

Solve for #dy/dx:#

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

Write everything in terms of #x:#

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

Answer 2 · 2018-04-02T00:06:55

Read below.

A bit of formula here:

#d/dx[f(x)^(g(x))]=f(x)^(g(x))*d/dx[ln(f(x))*g(x)]#

#=>d/dx[x^(1/x)]=x^(1/x)*d/dx[ln(x)*1/x]#

#d/dx[f(x)*g(x)]=f'(x)*g(x)+f(x)*g'(x)#

#=>d/dx[x^(1/x)]=x^(1/x)*(d/dx[ln(x)]*1/x+ln(x)*d/dx[1/x])#

Some rules here:

#d/dx[ln(x)]=1/x#

#d/dx[x^n]=nx^(n-1)# if #n# is a constant.

#=>d/dx[x^(1/x)]=x^(1/x)*(1/x*1/x+ln(x)*-1*x^(-1-1))#

#=>d/dx[x^(1/x)]=x^(1/x)(1/(x^2)+ln(x) *-1/(x^2))#

#=>d/dx[x^(1/x)]=x^(1/x)[1/(x^2)(1-ln(x))]#

#=>d/dx[x^(1/x)]=x^(1/x)[(1-ln(x))/(x^2)]#