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How do you differentiate #f(x)= 1/ (lnx)# using the quotient rule?

Answer 1

Isaiah Allen

#f'(x)=-1/(xln^2(x))#

As per the Quotient Rule,

Given

#f(x)=(h(x))/g(x)#

#:.f'(x)=(h'(x)*g(x)-g'(x)*h(x))/[g(x)]^2#

Regarding the query:

#h(x)=1=>h'(x)=0# #g(x)=ln(x)=>g'(x)=1/x#

Changing values in the formula:

#f'(x)=(0*lnx-1/x*1)/[lnx]^2=-(1/x)/[lnx]^2=-1/(xln^2(x))#

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Answer 2

Emily Ammerman

To differentiate the function ( f(x) = \frac{1}{\ln(x)} ) using the quotient rule:

[ f'(x) = \frac{u'v - uv'}{v^2} ]

Where: [ u = 1 ] [ v = \ln(x) ]

[ u' = 0 ] [ v' = \frac{1}{x} ]

Therefore,

[ f'(x) = \frac{0 \cdot \ln(x) - 1 \cdot \frac{1}{x}}{(\ln(x))^2} ] [ f'(x) = \frac{-1/x}{(\ln(x))^2} ] [ f'(x) = -\frac{1}{x \cdot (\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.