Home > Calculus > Differentiating Exponential Functions with Calculators

How do you differentiate #ln(1/x)#?

Answer 1

Elijah Abram

It is

#ln(1/x)=-lnx#

hence #d(ln(1/x))/dx=d(-lnx)/dx=-(d(lnx))/dx=-1/x#

Note we used the following identity

#ln(a/b)=lna-lnb#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Elijah Abram

#-1/x#

Instead of breaking #ln(1/x)# into #ln(1)-ln(x)=-ln(x)#, we can also use the rule that

#ln(a^b)=b*ln(a)#,

thus

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

Since #d/dx(ln(x))=1/x#, we see that #d/dx(-ln(x))=-1/x#.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 3

Christopher Agresta

To differentiate ln(1/x), use the chain rule and simplify the expression as follows:

d/dx ln(1/x) = d/dx ln(x^(-1)) = d/dx (-ln(x)) = -1/x.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 4

Charles Anctil

To differentiate ( \ln(1/x) ), you would use the chain rule. First, you would rewrite the expression as ( \ln(x^{-1}) ). Then, you would differentiate using the chain rule, which states that the derivative of ( \ln(u) ) with respect to ( x ) is ( \frac{1}{u} \frac{du}{dx} ). Applying this rule, the derivative of ( \ln(1/x) ) is ( \frac{1}{1/x} \cdot (-1/x^2) ), which simplifies to ( -\frac{1}{x} \cdot (-1/x^2) ), or ( \frac{1}{x^2} ). Therefore, the derivative of ( \ln(1/x) ) with respect to ( x ) is ( \frac{1}{x^2} ).

By signing up, you agree to our Terms of Service and Privacy Policy

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.