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How do you differentiate #f(x)=ln(x)^x#?

Answer 1

Christopher Agresta

#f'(x) = ln^x(x)(1/ln(x)+ln(ln(x)))#

Using implicit differentiation , the chain rule , and the product rule ,

Let #y = ln^x(x)#

#=>ln(y) = ln(ln^x(x)) = xln(ln(x))#

#=>d/dxln(y) = d/dxxln(ln(x))#

#=>1/ydy/dx = x(d/dxln(ln(x))) + ln(ln(x))(d/dxx)#

#=x(1/(ln(x))(d/dxln(x))) + ln(ln(x))#

#=x(1/ln(x)(1/x)) + ln(ln(x))#

#= 1/ln(x) + ln(ln(x))#

#=> dy/dx = y(1/ln(x) + ln(ln(x)))#

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

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Answer 2

Adam Adkins

To differentiate the function ( f(x) = \ln(x)^x ), you can use logarithmic differentiation.

Take the natural logarithm of both sides of the equation: ( \ln(f(x)) = \ln(\ln(x)^x) )
Apply the properties of logarithms to simplify the expression: ( \ln(f(x)) = x \ln(\ln(x)) )
Differentiate both sides of the equation with respect to ( x ): ( \frac{1}{f(x)} \cdot f'(x) = 1 \cdot \ln(\ln(x)) + x \cdot \frac{1}{\ln(x)} \cdot \frac{1}{x} )
Solve for ( f'(x) ): ( f'(x) = f(x) \left( \ln(\ln(x)) + \frac{1}{\ln(x)} \right) )
Substitute back the expression for ( f(x) ): ( f'(x) = \ln(x)^x \left( \ln(\ln(x)) + \frac{1}{\ln(x)} \right) )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.