What is the derivative of #ln(x)^x#?

Answer 1

We're going to use implicit differentiation to solve this problem.

We'll also be using the product rule and the chain rule .

Know that:

If #q=u*v# and #u=g(x)# and #v=h(x)#,
#(dq)/(dx)=u*(dv)/(dx)+v*(du)/(dx)#

So, say that:

#q=x*ln(ln(x))=u*v#

Therefore:

#u=x#, #(du)/(dx)=1#

Say that:

#v=ln(ln(x))=lnp#

Therefore:

#(dv)/(dp)=1/p=1/ln(x)#
#p=ln(x)#, #(dp)/(dx)=1/x#

This means that:

#(dv)/(dx)=1/(x*ln(x))#

And as a result...

#(dq)/(dx)=x*1/(x*ln(x))+ln(ln(x))*1#
#=1/(ln(x))+ln(ln(x))#

Now let's differentiate the function you were talking about using implicit differentiation...

#y=ln(x)^x#
#lny=ln(ln(x)^x)#
#lny=xln(ln(x))#
#1/y*(dy)/(dx)=1/(ln(x))+ln(ln(x))#
#y*1/y*(dy)/(dx)=y*{1/(ln(x))+ln(ln(x))}#
#(dy)/(dx)=y/(ln(x))+yln(ln(x))#

And finally...

#(dy)/(dx)=(ln(x)^x)/(ln(x))+ln(x)^x*ln(ln(x))#
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Answer 2

The derivative of ( \ln(x)^x ) with respect to ( x ) is given by:

[ \frac{d}{dx} (\ln(x)^x) = x\ln(x)^{x-1}\left(1 + \ln(\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.

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.

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.

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.

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