How do you differentiate #(x(lnx)^(x-1))(1/x)#?

Answer 1

I would first rewrite the function:

#(x(lnx)^(x-1))(1/x) = (lnx)^(x-1)#

Now use logarithmic differentiation:

#y = (lnx)^(x-1)#, so
#ln[y] = ln[ (lnx)^(x-1)]#, so
#lny = (x-1) ln(lnx)#.
Now differentiate implicitly (as we do for logarithmic differentiation). I use the product rule in the form: #d/(dx)(fg)=f'g+fg'#
Since, #g(x)=ln(lnx)# we have #g'(x)= 1/ lnx *1/x#
#1/y y' = (1) ln(lnx)+(x-1)[1/ lnx * 1/x]#
#y' = y[ln(lnx)+(x-1)/(x lnx)] = (lnx)^(x-1) [ln(lnx)+(x-1)/(x lnx)]#
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Answer 2

To differentiate the given expression (x(\ln(x))^{x-1}\cdot \frac{1}{x}), you can use the product rule and the chain rule.

Let's break it down:

  1. Product Rule: ((fg)' = f'g + fg')
  2. Chain Rule: ((g(h(x)))' = g'(h(x)) \cdot h'(x))

Now, let (f(x) = x) and (g(x) = (\ln(x))^{x-1}). Then, applying the product rule:

[f'(x) = 1] [g'(x) = (x-1)(\ln(x))^{x-2} \cdot \frac{1}{x}]

Now, applying the chain rule:

[g'(x) = (x-1)e^{(x-2)\ln(\ln(x))} \cdot \frac{1}{x}]

Now, let's put it all together:

[f'g = 1\cdot (\ln(x))^{x-1}] [fg' = x(\ln(x))^{x-1} \cdot (x-1)e^{(x-2)\ln(\ln(x))} \cdot \frac{1}{x}]

Therefore, the derivative of the given expression is:

[x(\ln(x))^{x-1} \cdot \frac{1}{x} + x(\ln(x))^{x-1} \cdot (x-1)e^{(x-2)\ln(\ln(x))} \cdot \frac{1}{x}]

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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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