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How do you differentiate #(lnx)^(x)#?

Answer 1

Emilia Aguilar

# d/dx(lnx)^x = (lnx)^x{1/lnx + ln((lnx))} #

Let #y=(lnx)^x#

Take (Natural) logarithms of both sided:

# " " lny = ln((lnx)^x ) # # :. lny = xln((lnx) ) #

Differentiate Implicitly (LHS) and apply product rule and chain rule (RHS).

# \ \ \ \ \ \ 1/ydy/dx = (x)(1/lnx*1/x) + (1)(ln((lnx)) # # :. \ 1/ydy/dx = 1/lnx + ln((lnx)) # # :. " " dy/dx = y{1/lnx + ln((lnx))} # # :. " " dy/dx = (lnx)^x{1/lnx + ln((lnx))} #

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Answer 2

Adam Adkins

To differentiate the function (ln(x))^x, you can use the chain rule and the logarithmic differentiation technique.

First, take the natural logarithm of the function:

y = (ln(x))^x ln(y) = x * ln(ln(x))

Now, differentiate both sides with respect to x:

(1/y) * dy/dx = ln(ln(x)) + x * (1/ln(x)) * (1/x)

Now, solve for dy/dx:

dy/dx = y * (ln(ln(x)) + 1/ln(x))

Substitute the expression for y:

dy/dx = (ln(x))^x * (ln(ln(x)) + 1/ln(x))

This is the derivative of the function (ln(x))^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.