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

2 Answers
Nov 18, 2017

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)#

Nov 18, 2017

#(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#