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

Answer 1

#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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Answer 2

To find the derivative of ( (\ln(x))^x ), we can use the chain rule. Let ( y = (\ln(x))^x ). Taking the natural logarithm of both sides, we get:

[ \ln(y) = x \ln(\ln(x)) ]

Now, differentiate both sides with respect to ( x ):

[ \frac{1}{y} \cdot \frac{dy}{dx} = \ln(\ln(x)) + \frac{x}{\ln(x)} \cdot \frac{1}{x} ]

[ \frac{dy}{dx} = y \left( \ln(\ln(x)) + \frac{1}{\ln(x)} \right) ]

Substitute back ( y = (\ln(x))^x ):

[ \frac{dy}{dx} = (\ln(x))^x \left( \ln(\ln(x)) + \frac{1}{\ln(x)} \right) ]

So, the derivative of ( (\ln(x))^x ) with respect to ( x ) is ( (\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.

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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