Home > Calculus > Differentiating Logarithmic Functions with Base e

How do you differentiate #y=1/lnx#?

Answer 1

William Abdalla

#=- 1/(x (ln x)^{2} )#

you can do this simply as #( (ln x)^{-1})'#

#=- (ln x)^{-2} (ln x)'#

#=- (ln x)^{-2} 1/x#

#=- 1/(x (ln x)^{2} )#

if you want to fiddle about with e and logs i suppose you could say that

#1/y = ln x#

#e^(1/y) = e^ln x = x#

so #(e^(1/y))' = 1#

and #( e^(1/y))' = e^(1/y) (1/y)'#

#= e^(1/y) * -(1/y^2) y'#

So #- e^(1/y) (1/y^2) y' = 1#

# y' = -y^2 * 1 / e^(1/y)#

# = -(1/ln x)^2 * 1/x#

#=- 1/(x (ln x)^{2} )#

same but bit more involved and fiddly

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Answer 2

Scarlett Allison

To differentiate ( y = \frac{1}{\ln(x)} ), you can use the chain rule. The derivative of ( \ln(x) ) is ( \frac{1}{x} ). Applying the chain rule, the derivative of ( \frac{1}{\ln(x)} ) is ( -\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.