Question

What is the derivative of `10log_4x/x`?

Answer 1

Derivative is `(10-10lnx)/(x^2ln4)`

Explanation:

As `log_4x=lnx/ln4`, let us find the derivative of `10/ln4*lnx/x`, using quotient formula for `f(x)=(g(x))/(h(x))`

`(df)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2`

Hence `d/(dx)10/ln4*lnx/x`

= `10/ln4*(1/x*x-1xxlnx)/x^2`

= `10/ln4*(1-lnx)/x^2`

= `(10-10lnx)/(x^2ln4)`

Answer 2

`(10(1/(ln4)-log_4x))/x^2`

Explanation:

`"differentiate using the "color(blue)"quotient rule"`

`"given "y=(g(x))/(h(x))" then "`

`dy/dx=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larr"quotient rule"`

`g(x)=10log_4xrArrg'(x)=10xx1/(xln4)`

`h(x)=xrArrh'(x)=1`

`rArrdy/dx=(x .10/(xln4)-10log_4x)/(x^2)`

`color(white)(rArrdy/dx)=(10(1/(ln4)-log_4x))/x^2`