Differentiate: `y = 2*x*log_4x` Answer should be: `y' = (1 + lnx)(1)/(ln2)`

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The derivative of `y=2x*log_4x` has to be determined.

`y = 2*x*log_4x`

=> `y = (2*x)*(ln x/ln 4)`

=> `y = 2*x*(ln x)/(ln 2^2)`

=> `y= (2*x*ln x)/(2*ln 2)`

=> `y = (1/ln 2)*x*ln x`

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

= `(1/ln 2)*(1*ln x + x*(1/x))`

= `(1/ ln 2)*(ln x +1)`

The derivative of `y= 2x*log_4 x` is `y'=(1/ln 2)*(1+ln x)`

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