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Category: Differentiation

use-the-first-principle-y-ln-cos-x-

Question Number 43854 by peter frank last updated on 16/Sep/18 $${use}\:{the}\:{first}\:{principle}\:{y}=\mathrm{ln}\:\sqrt{\mathrm{cos}\:{x}} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 16/Sep/18 $${y}+\bigtriangleup{y}={ln}\sqrt{{cos}\left({x}+\bigtriangleup{x}\right)}\:\: \\ $$$$\bigtriangleup{y}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left({cos}\left({x}+\bigtriangleup{x}\right)\right\}−\frac{\mathrm{1}}{\mathrm{2}}\left\{{lncosx}\right\}\right. \\ $$$$\bigtriangleup{y}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\frac{{cos}\left({x}+\bigtriangleup{x}\right)}{{cosx}}\right\}…

Q-by-using-fourier-series-prove-that-n-1-1-n-1-2n-1-3-3-32-

Question Number 174766 by mnjuly1970 last updated on 10/Aug/22 $$ \\ $$$$\:\:\:\:{Q}:\:\:{by}\:{using}\:{fourier}\:{series}, \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\Omega}=\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\:\boldsymbol{{n}}−\mathrm{1}} }{\left(\mathrm{2}\boldsymbol{{n}}−\mathrm{1}\:\right)^{\:\mathrm{3}} }\:=\:\frac{\boldsymbol{\pi}^{\:\mathrm{3}} }{\mathrm{32}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−−−−−− \\ $$…

Question-43438

Question Number 43438 by peter frank last updated on 10/Sep/18 Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18 $${cosh}^{−} \left(\mathrm{3}{x}\right)={ln}\left(\mathrm{3}{x}+\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}}\:\right) \\ $$$${y}=\left\{{ln}\left(\mathrm{3}{x}+\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{1}}\:\right)\right\}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}\left\{{ln}\left(\mathrm{3}{x}+\sqrt{\mathrm{9}{x}^{\mathrm{2}}…

Question-174442

Question Number 174442 by oustmuchiya@gmail.com last updated on 01/Aug/22 Answered by floor(10²Eta[1]) last updated on 01/Aug/22 $$\left.\mathrm{a}\right)\: \\ $$$$\mathrm{y}'=\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{4}} }{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}}+\mathrm{4}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} \sqrt{\mathrm{x}−\mathrm{1}} \\ $$$$\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} \left[\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}}+\mathrm{4}\sqrt{\mathrm{x}−\mathrm{1}}\right] \\…