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How do you find the derivative of #1/logx#?

Answer 1

Charles Arocha

It is #=-1/(xlog(x)^2)#.

With the chain rule.

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

So you apply the chain rule to the power and then to the log:

#d/dxlog(x)^-1=-1*log(x)^-2*d/dxlog(x)#

#=-1/(log(x)^2) 1/x#

#=-1/(xlog(x)^2)#.

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Answer 2

William Abdalla

To find the derivative of ( \frac{1}{\log(x)} ), use the chain rule. First, differentiate the outer function ( \frac{1}{u} ), where ( u = \log(x) ), with respect to ( u ), and then multiply by the derivative of ( u = \log(x) ) with respect to ( x ).

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

Therefore, the derivative of ( \frac{1}{\log(x)} ) is ( -\frac{1}{x(\log(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.