Question Number 11580 by Nayon last updated on 28/Mar/17 $${why}\:\frac{{dy}}{{dx}}=\frac{{dy}}{{du}}.\frac{{du}}{{dx}} \\ $$ Answered by mrW1 last updated on 28/Mar/17 $${the}\:{chain}\:{rule}\:{is}\:{one}\:{of}\:{the}\:{elementary} \\ $$$${rules}.\:{you}\:{should}\:{know}\:{them},\:{but}\:\: \\ $$$${you}\:{don}'{t}\:{need}\:{to}\:{prove}\:{them}.\:{if} \\…
Question Number 11571 by Nayon last updated on 28/Mar/17 $${why}\:\:\:\frac{{d}\left[{f}\left\{{g}\left({x}\right)\right\}\right]}{{dx}}=\frac{{df}\left[\left\{{g}\left({x}\right)\right\}\right]}{{dg}\left({x}\right)}.\frac{{dg}\left({x}\right)}{{dx}}? \\ $$ Answered by mrW1 last updated on 28/Mar/17 $${y}={f}\left({u}\right) \\ $$$${u}={g}\left({x}\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{du}}×\frac{{du}}{{dx}} \\…
Question Number 11567 by Nayon last updated on 28/Mar/17 $${why}\:\:\:\:\:\:{li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\:\frac{{f}\left({x}\right)}{{g}\left({x}\right)}={li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{f}'\left({x}\right)}{{g}'\left({x}\right)} \\ $$ Answered by mrW1 last updated on 28/Mar/17 $${l}'{hopital}'{s}\:{rule} \\ $$ Terms…
Question Number 11565 by Nayon last updated on 28/Mar/17 $${if}\:\frac{{dy}}{{dx}}={p}\:\:,{then}\:{why}\:{dy}={pdx}? \\ $$ Answered by mrW1 last updated on 28/Mar/17 $${if}\:\frac{{dy}}{{dx}}={p}\Rightarrow{y}={px} \\ $$$$\Rightarrow{dy}={pdx} \\ $$ Terms…
Question Number 11563 by Nayon last updated on 28/Mar/17 $${if}\:\:{f}\left({x}\right)={g}\left({y}\right) \\ $$$${then}\:{why}\:\frac{{d}}{{dx}}\left({f}\left({x}\right)\right)=\frac{{d}}{{dy}}\left({g}\left({y}\right)\right)? \\ $$ Commented by mrW1 last updated on 28/Mar/17 $${this}\:{is}\:{not}\:{always}\:{correct}.\:{but} \\ $$$$\frac{{d}}{{dx}}\left({f}\left({x}\right)\right)=\frac{{d}}{{dy}}\left({g}\left({y}\right)\right)×\frac{{dy}}{{dx}} \\…
Question Number 11552 by Nayon last updated on 28/Mar/17 $${find}\:\frac{{d}}{{dx}}\left({y}\right)\:{where}\:\:{y}=\overset{\sqrt{{x}}} {\:}\sqrt{\sqrt{{x}}} \\ $$$$ \\ $$ Answered by ajfour last updated on 28/Mar/17 $${y}\:=\:\left(\sqrt{{x}}\right)^{\frac{\mathrm{1}}{\:\sqrt{{x}}}} \\ $$$$\mathrm{ln}\:{y}\:=\frac{\mathrm{ln}\:\sqrt{{x}}}{\:\sqrt{{x}}}…
Question Number 11547 by Nayon last updated on 28/Mar/17 $$\:\:\:\:\:\:\:\:\:\:\frac{{d}}{{dx}}\left({x}^{{x}} \right)=?\left[{please}\:{give}\:{the}\:{answer}\:{with}\:{proof}\right] \\ $$ Answered by sma3l2996 last updated on 28/Mar/17 $${x}^{{x}} ={e}^{{xln}\left({x}\right)} \\ $$$${so}\:\:\frac{{d}\left({x}^{{x}} \right)}{{dx}}=\frac{{d}\left({e}^{{xln}\left({x}\right)}…
Question Number 11548 by Nayon last updated on 28/Mar/17 $$ \\ $$$${find}\:\frac{{d}}{{dx}}\left(\mathrm{2}^{{x}} \right) \\ $$ Answered by sma3l2996 last updated on 28/Mar/17 $$=\frac{{d}\left({e}^{{xln}\left(\mathrm{2}\right)} \right)}{{dx}}={ln}\left(\mathrm{2}\right){e}^{{xln}\left(\mathrm{2}\right)} ={ln}\left(\mathrm{2}\right)\mathrm{2}^{{x}}…
Question Number 11545 by Nayon last updated on 28/Mar/17 $${find}\frac{{dy}}{{dx}}\:{if}\:{x}^{{x}} {y}^{{y}} =\mathrm{1} \\ $$ Answered by Joel576 last updated on 28/Mar/17 $${y}^{{y}} \:=\:{x}^{−{x}} \\ $$$${y}\:\mathrm{ln}\:{y}\:=\:−{x}\:\mathrm{ln}\:{x}…
Question Number 11546 by Nayon last updated on 28/Mar/17 $$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:{f}\left({x}\right)=^{{x}} \sqrt{{x}}\:{find}\:\frac{{d}}{{dx}}\left({f}\left({x}\right)\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{and}\:{for}\:{what}\:{x}\:,{we}\:{will}\:{get}\:{the}\: \\ $$$$\:{maximum}\:{of}\:{the}\:{function}..? \\ $$$$…