Hi! could you please determine the derivative of the following function? `y=log_x(3)` thanks

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Correction:

`y'=-ln(3)*1/(ln(x))^2 *1/x = -(ln(3))/(x(lnx)^2)`

Or

`y'=-(log_x(3))/(xlnx)`

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First, simplify the given function using the change of base formula for logarithms:

`log_b (a) = (log_c (a))/(log_c (a))`

Applying this formula to the given function in order to change to logarithm with base e (natural log) , we get

`y = log_x(3) = (ln(3))/(ln(x))`

The derivaltive can now be found using the chain rule and the fact that `(ln(x))' = 1/x` :

`y' = ln(3) * 1/(ln(x))^2 * 1/x = (ln(3))/(x(ln(x))^2)`

We could also rewrite it back as a log with base x:

`y' = (log_x(3))/(xlnx)`

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