If f(x)=sqrt 23x, find f'(x)

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`f(x)=sqrt (23x)`

Express the square root as exponent.

`f(x)=(23x)^(1/2)`

To take the derivative of f(x), apply formula which is `(u^n)' = n*u^(n-1)* u'` .

`f'(x)=1/2*(23x)^(1/2 - 1)*(23x)' = 1/2*(23x)^(-1/2)*23 = (23(23x)^(-1/2))/2`

Apply, the negative property of exponent which is a^(-m)=1/a^m.

`f'(x)=23/(2(23x)^(1/2))`

Then, express the fractional exponent as radical.

`f'(x)=23/(2sqrt(23x))`

**Hence, the derivative of the given function is `f'(x)=23/(2sqrt(23x))` .**

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