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How do you differentiate #y = 1 / log_2 x#?

Answer 1

Emma Aldridge

#(dy)/(dx)=-ln2/(x(lnx)^2)#

#log_2x=lnx/ln2#

Therefore #y=ln2/lnx=ln2*1/lnx#

and #(dy)/(dx)=ln2*(-1/(lnx)^2)*1/x#

= #-ln2/(x(lnx)^2)#

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Answer 2

Adam Adkins

To differentiate ( y = \frac{1}{\log_2 x} ) with respect to ( x ), we'll use the chain rule and the derivative of the natural logarithm.

First, rewrite the expression as ( y = \frac{1}{\ln x} \div \frac{1}{\ln 2} ).

Then, differentiate ( y ) with respect to ( x ) using the chain rule:

( \frac{dy}{dx} = -\frac{1}{{(\ln x)}^2} \times \frac{1}{x} )

Finally, substitute ( \ln 2 ) for ( \ln x ):

( \frac{dy}{dx} = -\frac{1}{{(\ln x)}^2} \times \frac{1}{x} \times \frac{1}{\ln 2} )

So, the derivative of ( y ) with respect to ( x ) is:

( \frac{dy}{dx} = -\frac{1}{{x \cdot (\ln x)}^2 \cdot \ln 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.