The-derivative-of-F-x-x-2-x-3-1-log-t-dt-x-gt-0-is-

Question Number 53532 by gunawan last updated on 23/Jan/19

The derivative of F (x)=∫_x^2 ^x^3 (1/(log t)) dt (x >0) is

$$\mathrm{The}\:\mathrm{derivative}\:\mathrm{of}\:{F}\:\left({x}\right)=\underset{{x}^{\mathrm{2}} } {\overset{{x}^{\mathrm{3}} } {\int}}\:\frac{\mathrm{1}}{\mathrm{log}\:{t}}\:{dt} \\ $$$$\left({x}\:>\mathrm{0}\right)\:\mathrm{is} \\ $$

Commented by Abdo msup. last updated on 23/Jan/19

(dF/dx)(x)=(((x^3 )^′ )/(log(x^3 ))) −(((x^2 )^′ )/(log(x^2 ))) =((3x^2 )/(3log(x))) −((2x)/(2log(x))) =((x^2 −x)/(log(x)))

$$\frac{{dF}}{{dx}}\left({x}\right)=\frac{\left({x}^{\mathrm{3}} \right)^{'} }{{log}\left({x}^{\mathrm{3}} \right)}\:−\frac{\left({x}^{\mathrm{2}} \right)^{'} }{{log}\left({x}^{\mathrm{2}} \right)}\:=\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{3}{log}\left({x}\right)}\:−\frac{\mathrm{2}{x}}{\mathrm{2}{log}\left({x}\right)} \\ $$$$=\frac{{x}^{\mathrm{2}} −{x}}{{log}\left({x}\right)} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jan/19

((dF(x))/dx)=∫_x^2 ^x^3 (∂/∂x)((1/(logt)))dt +(1/(logx^3 ))×(d/dx)(x^3 )−(1/(logx^2 ))×(d/dx)(x^2 ) =0+((3x^2 )/(3logx))−((2x)/(2logx)) =(x^2 /(logx))−(x/(logx))=(x/(logx))(x−1)

$$\frac{{dF}\left({x}\right)}{{dx}}=\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} } \:\frac{\partial}{\partial{x}}\left(\frac{\mathrm{1}}{{logt}}\right){dt}\:+\frac{\mathrm{1}}{{logx}^{\mathrm{3}} }×\frac{{d}}{{dx}}\left({x}^{\mathrm{3}} \right)−\frac{\mathrm{1}}{{logx}^{\mathrm{2}} }×\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} \right) \\ $$$$=\mathrm{0}+\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{3}{logx}}−\frac{\mathrm{2}{x}}{\mathrm{2}{logx}} \\ $$$$=\frac{{x}^{\mathrm{2}} }{{logx}}−\frac{{x}}{{logx}}=\frac{{x}}{{logx}}\left({x}−\mathrm{1}\right) \\ $$

Commented by gunawan last updated on 23/Jan/19