Question Number 19903 by j.masanja06@gmail.com last updated on 17/Aug/17 $$\mathrm{by}\:\mathrm{use}\:\mathrm{the}\:\mathrm{first}\: \\ $$$$\mathrm{principle},\mathrm{find} \\ $$$$\mathrm{dy}/\mathrm{dx}\:\mathrm{of}\: \\ $$$$\mathrm{y}=\mathrm{cos}\left(\mathrm{x}−\frac{\Pi}{\mathrm{8}}\right) \\ $$ Answered by icyfalcon999 last updated on 18/Aug/17…
Question Number 150963 by EDWIN88 last updated on 17/Aug/21 $${Given}\:{x}\:,{y}\:{real}\:{number}\:{such}\:{that} \\ $$$$\:\mathrm{0}<\frac{{y}}{{x}}<\frac{\mathrm{1}}{\mathrm{2}}.\:{Find}\:{minimum}\:{value} \\ $$$${of}\:\frac{\mathrm{2}{y}}{{x}−{y}}\:+\frac{\mathrm{3}{x}}{{x}+\mathrm{2}{y}}\:.\: \\ $$ Answered by john_santu last updated on 17/Aug/21 $$\:\mathrm{let}\:\frac{\mathrm{y}}{\mathrm{x}}\:=\:\mathrm{t}\:\Rightarrow\mathrm{f}\left(\mathrm{t}\right)=\frac{\mathrm{2t}}{\mathrm{1}−\mathrm{t}}+\frac{\mathrm{3}}{\mathrm{1}+\mathrm{2t}}\: \\…
Question Number 19859 by vivek last updated on 16/Aug/17 $$\frac{\sqrt{}\mathrm{1}+{x}−\sqrt{}\mathrm{1}−{x}}{\:\sqrt{}\mathrm{1}+{x}−\sqrt{}\mathrm{1}−{x}}\:\:\boldsymbol{{differentiate}}\:\boldsymbol{{w}}.\boldsymbol{{r}}.\boldsymbol{{t}}\:\boldsymbol{{x}} \\ $$ Commented by giridhar.mathmania last updated on 17/Aug/17 $$\mathrm{function}\:\mathrm{to}\:\mathrm{be}\:\mathrm{differentiated}\:\mathrm{is}\:\mathrm{1}. \\ $$$$\mathrm{hence}\:\mathrm{its}\:\mathrm{differential}\:\mathrm{is}\:\mathrm{0}. \\ $$ Terms…
Question Number 19860 by vivek last updated on 16/Aug/17 $$\mathrm{log}_{{e}} \sqrt{\frac{\mathrm{1}−{cox}}{\mathrm{1}+{cosx}}\:\:\boldsymbol{{differentiate}}\:\boldsymbol{{w}}.\boldsymbol{{r}}.\boldsymbol{{t}}\:\boldsymbol{{x}}} \\ $$ Answered by myintkhaing last updated on 17/Aug/17 $$\frac{\mathrm{d}}{\mathrm{dx}}\:\left(\mathrm{log}_{\mathrm{e}} \:\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\right) \\ $$$$=\:\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{x}}\right)\left(\frac{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\left(−\mathrm{sin}\:\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} }\right)…
Question Number 150828 by mnjuly1970 last updated on 15/Aug/21 $$ \\ $$$$\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{sin}^{\:\mathrm{2}} \left({x}\:\right)}{{x}\sqrt{{x}}}\:{dx}\overset{?} {=}\:\sqrt{\pi} \\ $$ Answered by puissant last updated on 16/Aug/21…
Question Number 85248 by jagoll last updated on 20/Mar/20 $$\mathrm{find}\:\mathrm{minimum} \\ $$$$\mathrm{y}\:=\:\mathrm{2}\mid\mathrm{x}−\mathrm{3}\mid\:+\:\mid\mathrm{4x}+\mathrm{2}\mid\: \\ $$ Commented by mr W last updated on 20/Mar/20 $${x}−\mathrm{3}=\mathrm{0}\:\Rightarrow{x}=\mathrm{3} \\ $$$$\mathrm{4}{x}+\mathrm{2}=\mathrm{0}\:\Rightarrow{x}=−\frac{\mathrm{1}}{\mathrm{2}}…
Question Number 150749 by mnjuly1970 last updated on 15/Aug/21 $$ \\ $$$$\:\:\mathrm{Prove}\:\:\:\mathrm{That}\::: \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathcal{I}\::=\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\left(\:{sin}\:\left({x}\:\right).{cos}\:\left({x}\:\right)\right)^{\:\mathrm{4}} }{{x}\:.\:\sqrt{{x}}\:}{dx}=\frac{\mathrm{1}}{\mathrm{32}}\:\left(\mathrm{2}\:\sqrt{\mathrm{2}}\:−\mathrm{1}\:\right)\sqrt{\:\pi}\:….\blacksquare\:\: \\ $$$$\:\:\:..{m}.{n}..\:\: \\ $$ Terms of…
Question Number 150747 by mnjuly1970 last updated on 15/Aug/21 $$ \\ $$$${prove}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega:=\int_{\mathrm{0}} ^{\:\infty} {e}^{−\sqrt{{x}}} .{ln}\:\left(\sqrt[{\mathrm{4}}]{\:{x}}\:\right)\overset{?} {=}\:\mathrm{1}−\gamma \\ $$$$\:{m}.{n}.. \\ $$ Answered by Olaf_Thorendsen…
Question Number 19657 by vivek last updated on 14/Aug/17 $${differentiate}\:{the}\:{function}\:{with}\:{respect}\: \\ $$$${to}\:{x} \\ $$$$\mathrm{1}.\:\:\:\:\frac{{secx}−\mathrm{1}}{{secx}+\mathrm{1}} \\ $$ Commented by prakash jain last updated on 14/Aug/17 $${answer}\:{in}\:{Q}\mathrm{19658}…
Question Number 85169 by jagoll last updated on 19/Mar/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{n}^{\mathrm{th}} \:\mathrm{derivative}\:\mathrm{of}\:\mathrm{function} \\ $$$$\mathrm{y}\:=\:\sqrt{\mathrm{sin}\:\mathrm{x}}\:\mathrm{by}\:\mathrm{Leibniz}\:\mathrm{theorem} \\ $$ Commented by mr W last updated on 19/Mar/20 $${i}\:{don}'{t}\:{think}\:{Leibniz}\:{theorem}\:{helps} \\…