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How do you differentiate #y=ln(1/x)-1/lnx#?

Answer 1

Aurora Alvarado

#y'=-1/x+1/(x(lnx)^2)#

Let's rewrite the function a bit, mostly to get rid of #ln(1/x).#

#y=(ln1-lnx)-(lnx)^-1#

Since #ln1=0:#

#y=-lnx-(lnx)^-1#

#y'=-1/x-(-(lnx)^-2*d/dxlnx)# (Chain Rule)

#y'=-1/x+1/(x(lnx)^2)#

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Answer 2

Ezra Adams

To differentiate ( y = \ln(1/x) - \frac{1}{\ln(x)} ), you can apply the chain rule and the derivative of natural logarithm function.

[ y' = \frac{d}{dx}\left(\ln(1/x)\right) - \frac{d}{dx}\left(\frac{1}{\ln(x)}\right) ]

Apply the chain rule for both terms:

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

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

Substitute these derivatives back into the equation:

[ y' = -\frac{1}{x} + \frac{1}{x(\ln(x))^2} ]

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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.