# How do you differentiate #f(x)= sqrt (ln2x)/x#?

By using the quotient rule, together with the power rule, we get

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To differentiate ( f(x) = \frac{\sqrt{\ln(2x)}}{x} ), we can use the quotient rule, which states that if we have a function of the form ( \frac{u(x)}{v(x)} ), then its derivative is given by ( \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} ).

First, we identify ( u(x) = \sqrt{\ln(2x)} ) and ( v(x) = x ).

Next, we find the derivatives of ( u(x) ) and ( v(x) ):

[ u'(x) = \frac{1}{2\sqrt{\ln(2x)}} \cdot \frac{1}{2x} \cdot 2 = \frac{1}{2x\sqrt{\ln(2x)}} ]

[ v'(x) = 1 ]

Now, we apply the quotient rule:

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

[ f'(x) = \frac{1}{2x\sqrt{\ln(2x)}} - \frac{\sqrt{\ln(2x)}}{x^2} ]

Simplifying, we get:

[ f'(x) = \frac{1}{2x\sqrt{\ln(2x)}} - \frac{\sqrt{\ln(2x)}}{x^2} ]

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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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