# What is the derivative of #y=log_10x/x#?

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To find the derivative of ( y = \frac{\log_{10} x}{x} ), you can use the quotient rule.

Let ( u = \log_{10} x ) and ( v = x ).

Then, ( u' = \frac{1}{x \ln 10} ) and ( v' = 1 ).

Using the quotient rule:

[ \frac{dy}{dx} = \frac{u'v - uv'}{v^2} ]

[ = \frac{\frac{1}{x \ln 10} \cdot x - \log_{10} x \cdot 1}{x^2} ]

[ = \frac{1}{x \ln 10} - \frac{\log_{10} x}{x^2} ]

Therefore, the derivative of ( y = \frac{\log_{10} x}{x} ) is ( \frac{1}{x \ln 10} - \frac{\log_{10} x}{x^2} ).

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To find the derivative of ( y = \frac{\log_{10}x}{x} ), you can use the quotient rule:

Given ( y = \frac{\log_{10}x}{x} ),

The derivative is given by:

[ \frac{d}{dx}\left(\frac{\log_{10}x}{x}\right) = \frac{\frac{d}{dx}(\log_{10}x) \cdot x - \log_{10}x \cdot \frac{d}{dx}(x)}{x^2} ]

Using the chain rule for differentiation, (\frac{d}{dx}(\log_{10}x) = \frac{1}{x \ln(10)} ).

And (\frac{d}{dx}(x) = 1).

So, plugging into the formula, we get:

[ \frac{1}{x \ln(10)} \cdot x - \frac{\log_{10}x \cdot 1}{x^2} = \frac{1}{\ln(10)} - \frac{\log_{10}x}{x^2} ]

Thus, the derivative of ( y = \frac{\log_{10}x}{x} ) is ( \frac{1}{\ln(10)} - \frac{\log_{10}x}{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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