What is the derivative of #f(x)=ln(x)/x# ?

Answer 1

By Quotient Rule,

#y'={1/x cdot x-lnx cdot 1}/{x^2}={1-lnx}/{x^2}#

This problem can also be solved by the Product Rule

#y'=f'(x)g(x)+f(x)g(x)#

The original function can also be rewritten using negative exponents.

#f(x)=ln(x)/x=ln(x)*x^-1#
#f'(x)=1/x*x^-1+ln(x)*-1x^-2#
#f'(x)=1/x*1/x+ln(x)*-1/x^2#
#f'(x)=1/x^2-ln(x)/x^2#
#f'(x)=(1-ln(x))/x^2#
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Answer 2

To find the derivative of ( f(x) = \frac{\ln(x)}{x} ), you can use the quotient rule of differentiation. The quotient rule states that if you have a function ( u(x) ) divided by ( v(x) ), then the derivative is given by ( \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} ).

In this case: [ u(x) = \ln(x) ] [ v(x) = x ]

The derivatives are: [ u'(x) = \frac{1}{x} ] [ v'(x) = 1 ]

Now, applying the quotient rule: [ f'(x) = \frac{\frac{1}{x} \cdot x - \ln(x) \cdot 1}{x^2} ] [ f'(x) = \frac{1 - \ln(x)}{x^2} ]

So, the derivative of ( f(x) = \frac{\ln(x)}{x} ) is ( f'(x) = \frac{1 - \ln(x)}{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.

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.

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.

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.

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