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Category: Differentiation

f-x-ln-x-f-f-

Question Number 57754 by malwaan last updated on 11/Apr/19 $$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right) \\ $$$$\left(\boldsymbol{{f}}\circ\boldsymbol{{f}}\right)'=? \\ $$ Commented by maxmathsup by imad last updated on 11/Apr/19 $${we}\:{have}\:{f}^{'} \left({x}\right)\:=\frac{\mathrm{1}}{{x}}\:\Rightarrow\:\left({fof}\right)^{'}…

y-x-dy-dx-a-y-y-dy-dx-

Question Number 123282 by zarminaawan last updated on 24/Nov/20 $${y}−{x}\frac{{dy}}{{dx}}={a}\left({y}×{y}+\frac{{dy}}{{dx}}\right) \\ $$ Answered by Dwaipayan Shikari last updated on 24/Nov/20 $${y}−{x}\frac{{dy}}{{dx}}={ay}^{\mathrm{2}} +{a}\frac{{dy}}{{dx}} \\ $$$$\frac{{dy}}{{dx}}\left({x}+{a}\right)={y}\left(\mathrm{1}−{ay}\right) \\…

Question-123124

Question Number 123124 by aupo14 last updated on 23/Nov/20 Commented by Dwaipayan Shikari last updated on 23/Nov/20 $${x}!=\Gamma\left({x}+\mathrm{1}\right) \\ $$$${log}\left({x}!\right)={log}\left(\Gamma\left({x}+\mathrm{1}\right)\right) \\ $$$$\frac{\frac{{dx}!}{{dx}}}{{x}!}=\frac{\Gamma'\left({x}+\mathrm{1}\right)}{\Gamma\left({x}+\mathrm{1}\right)} \\ $$$$\Rightarrow\frac{{d}}{{dx}}\left({x}!\right)={x}!.\psi\left({x}+\mathrm{1}\right) \\…

Question-188534

Question Number 188534 by thotasandeep111 last updated on 03/Mar/23 Answered by mr W last updated on 03/Mar/23 $$\int{f}\left({x}\right){dx}={f}\left({x}\right) \\ $$$$\Rightarrow{f}\left({x}\right)={e}^{{x}} \\ $$$$\int\left({f}\left({x}\right)\right)^{\mathrm{2}} {dx}=\int{e}^{\mathrm{2}{x}} {dx}=\frac{{e}^{\mathrm{2}{x}} }{\mathrm{2}}+{C}=\frac{\left({f}\left({x}\right)\right)^{\mathrm{2}}…

solve-x-1-y-1-x-y-x-e-2x-

Question Number 57415 by Abdo msup. last updated on 03/Apr/19 $${solve}\:\left({x}−\mathrm{1}\right){y}^{'} \:+\left(\mathrm{1}+\sqrt{{x}}\right){y}\:={x}\:{e}^{−\mathrm{2}{x}} \\ $$ Commented by maxmathsup by imad last updated on 12/Apr/19 $${due}\:{to}\:\sqrt{{x}}{we}\:{must}\:{have}\:{x}\geqslant\mathrm{0}\:\:\:\left({ed}\right)\:\Leftrightarrow\left(\sqrt{{x}}−\mathrm{1}\right)\left(\sqrt{{x}}+\mathrm{1}\right){y}^{'} +\left(\sqrt{\:{x}}+\mathrm{1}\right){y}\:={xe}^{−\mathrm{2}{x}}…