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How do you find the derivative of #ln(10/x)#?

Answer 1

Luke Alvarez

# d/dxln(10/x) = -1/x #

Simply the expression using the rule of logs and use #d/dx(lnx)=1/x#

So, # d/dxln(10/x) = d/dx{ln10-lnx} #

# = d/dxln10-d/dxlnx # # = 0-1/x # # = -1/x #

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Answer 2

Charles Anctil

To find the derivative of ( \ln\left(\frac{10}{x}\right) ) with respect to (x), you apply the chain rule:

[ \frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx} ]

In this case, (u = \frac{10}{x}).

[ \frac{du}{dx} = \frac{d}{dx}\left(\frac{10}{x}\right) = -\frac{10}{x^2} ]

Therefore, the derivative of ( \ln\left(\frac{10}{x}\right) ) with respect to (x) is:

[ \frac{d}{dx}\left[\ln\left(\frac{10}{x}\right)\right] = \frac{1}{\frac{10}{x}} \cdot \left(-\frac{10}{x^2}\right) = -\frac{x}{10} \cdot \frac{-10}{x^2} = \frac{1}{x} ]

So, the derivative is ( \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.