Question Number 53532 by gunawan last updated on 23/Jan/19 | ||
$$\mathrm{The}\:\mathrm{derivative}\:\mathrm{of}\:{F}\:\left({x}\right)=\underset{{x}^{\mathrm{2}} } {\overset{{x}^{\mathrm{3}} } {\int}}\:\frac{\mathrm{1}}{\mathrm{log}\:{t}}\:{dt} \\ $$ $$\left({x}\:>\mathrm{0}\right)\:\mathrm{is} \\ $$ | ||
Commented byAbdo msup. last updated on 23/Jan/19 | ||
$$\frac{{dF}}{{dx}}\left({x}\right)=\frac{\left({x}^{\mathrm{3}} \right)^{'} }{{log}\left({x}^{\mathrm{3}} \right)}\:−\frac{\left({x}^{\mathrm{2}} \right)^{'} }{{log}\left({x}^{\mathrm{2}} \right)}\:=\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{3}{log}\left({x}\right)}\:−\frac{\mathrm{2}{x}}{\mathrm{2}{log}\left({x}\right)} \\ $$ $$=\frac{{x}^{\mathrm{2}} −{x}}{{log}\left({x}\right)} \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jan/19 | ||
$$\frac{{dF}\left({x}\right)}{{dx}}=\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} } \:\frac{\partial}{\partial{x}}\left(\frac{\mathrm{1}}{{logt}}\right){dt}\:+\frac{\mathrm{1}}{{logx}^{\mathrm{3}} }×\frac{{d}}{{dx}}\left({x}^{\mathrm{3}} \right)−\frac{\mathrm{1}}{{logx}^{\mathrm{2}} }×\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} \right) \\ $$ $$=\mathrm{0}+\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{3}{logx}}−\frac{\mathrm{2}{x}}{\mathrm{2}{logx}} \\ $$ $$=\frac{{x}^{\mathrm{2}} }{{logx}}−\frac{{x}}{{logx}}=\frac{{x}}{{logx}}\left({x}−\mathrm{1}\right) \\ $$ | ||
Commented bygunawan last updated on 23/Jan/19 | ||
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir} \\ $$ | ||
Commented bymalwaan last updated on 23/Jan/19 | ||
$$\mathrm{0}\:\:\mathrm{how}? \\ $$ | ||
Commented bytanmay.chaudhury50@gmail.com last updated on 23/Jan/19 | ||
$${partial}\:{differentiation}\:{of}\:{F}\left({t}\right)\:{w}.{r}.{t}.{x}\:{is}\:{x}\:{is}\:\mathrm{0} \\ $$ | ||
Commented bymalwaan last updated on 24/Jan/19 | ||
$$\mathrm{Yes} \\ $$ $$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir} \\ $$ | ||