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Category: Differentiation

lim-x-0-tan-3x-3tan-x-x-3-

Question Number 87378 by jagoll last updated on 04/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{3x}−\mathrm{3tan}\:\mathrm{x}}{\mathrm{x}^{\mathrm{3}} } \\ $$ Commented by john santu last updated on 04/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{3x}+\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{3x}\right)^{\mathrm{3}} +\mathrm{o}\left(\left(\mathrm{3x}\right)^{\mathrm{3}}…

Question-152904

Question Number 152904 by DELETED last updated on 03/Sep/21 Answered by DELETED last updated on 03/Sep/21 $$\left.\mathrm{1}\right).\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{4cos}\:\mathrm{x}+\mathrm{5sin}\:\mathrm{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:\:\:=−\mathrm{4}\:\mathrm{sin}\:\mathrm{x}+\mathrm{5}\:\mathrm{cos}\:× \\ $$$$\left.\mathrm{2}\right).\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{3}\:\mathrm{sin}\:\mathrm{2x}\:−\:\mathrm{5}\:\mathrm{cos}\:\mathrm{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:\:=\mathrm{3}×\mathrm{2}\:\mathrm{cos}\:\mathrm{2x}\:+\mathrm{5}\:\mathrm{sin}\:\mathrm{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{6}\:\mathrm{cos}\:\mathrm{2x}\:+\:\mathrm{5}\:\mathrm{sin}\:\mathrm{x}//…

nice-mathematics-Prove-that-I-0-cos-x-cosh-x-dx-pi-cosh-pi-2-prepared-m-n-

Question Number 152797 by mnjuly1970 last updated on 01/Sep/21 $$ \\ $$$$\:\:\:{nice}..{mathematics}… \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\mathrm{Prove}\:\mathrm{that}… \\ $$$$\:\mathrm{I}=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{cos}\:\left({x}\:\right)}{{cosh}\:\left({x}\:\right)}\:{dx}=\frac{\pi}{\:{cosh}\:\left(\frac{\pi}{\mathrm{2}}\:\right)}\:…….\blacksquare\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:{prepared}\:::\:\:{m}.{n} \\ $$$$ \\…

Question-152631

Question Number 152631 by mnjuly1970 last updated on 30/Aug/21 Answered by Olaf_Thorendsen last updated on 30/Aug/21 $$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} +{x}^{\mathrm{4}} \right)}{{x}^{\mathrm{2}} }\:{dx} \\ $$$$\mathrm{I}\:=\:\left[−\frac{\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} +{x}^{\mathrm{4}}…