Question Number 84063 by niroj last updated on 09/Mar/20 $$ \\ $$$$\:\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\theta}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{Mean}}\:\boldsymbol{\mathrm{Value}} \\ $$$$\:\boldsymbol{\mathrm{Theorem}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{h}}\right)=\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)+\boldsymbol{\mathrm{h}}\:\boldsymbol{\mathrm{f}}^{\:'} \:\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\theta\mathrm{h}}\right)\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 84059 by niroj last updated on 09/Mar/20 $$ \\ $$$$\:\boldsymbol{\mathrm{Find}}\:\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}\:\boldsymbol{\mathrm{of}}\:\:\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{m}}} \boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}} \:=\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\right)^{\boldsymbol{\mathrm{m}}+\boldsymbol{\mathrm{n}}} . \\ $$ Commented by jagoll last updated on 09/Mar/20 $${x}^{{m}+{n}}…
Question Number 18470 by tawa tawa last updated on 22/Jul/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{partial}\:\mathrm{derivatives}\:\mathrm{for}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Z}\:=\:\mathrm{3x}^{\mathrm{2}} \left(\mathrm{5x}\:+\:\mathrm{7y}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Z}\:=\:\left(\mathrm{w}\:−\:\mathrm{x}\:−\:\mathrm{y}\right)^{\mathrm{2}} \:\left(\mathrm{3w}\:+\:\mathrm{2x}\:−\:\mathrm{4y}\right) \\ $$ Answered by ajfour last updated on…
Question Number 18469 by tawa tawa last updated on 22/Jul/17 $$\mathrm{Find}\:\mathrm{Z}_{\mathrm{x}} \:\mathrm{and}\:\mathrm{Z}_{\mathrm{y}} \:\mathrm{for}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{functions}\:\mathrm{below} \\ $$$$\left(\mathrm{a}\right)\:\:\mathrm{Z}\:=\:\mathrm{8x}^{\mathrm{2}} \mathrm{y}\:+\:\mathrm{14xy}^{\mathrm{2}} \:+\:\mathrm{5y}^{\mathrm{2}} \mathrm{x}^{\mathrm{3}} \\ $$$$\left(\mathrm{b}\right)\:\:\mathrm{Z}\:=\:\mathrm{4x}^{\mathrm{3}} \mathrm{y}^{\mathrm{2}} \:+\:\mathrm{2x}^{\mathrm{2}} \mathrm{y}^{\mathrm{3}} \:−\:\mathrm{7xy}^{\mathrm{5}} \\…
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Question Number 18283 by tawa tawa last updated on 17/Jul/17 $$\mathrm{Given}:\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{Show}\:\mathrm{that},\:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=\:\frac{\mathrm{xy}}{\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{2}} } \\ $$ Terms of Service Privacy Policy…
Question Number 149339 by mnjuly1970 last updated on 04/Aug/21 $$\:\:\:\:\:\:\:\:\:\:…{nice}……{mathematics}… \\ $$$$\:\:\:\:\:\:{ln}\left(\:\mathrm{2}\right)\:−\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{\zeta\:\left(\:\mathrm{2}{n}+\mathrm{1}\:\right)−\mathrm{1}}{{n}\:+\:\mathrm{1}}\:=\:? \\ $$$$\:\:\:\:….{m}.{n}…. \\ $$ Answered by Kamel last updated on 04/Aug/21…
Question Number 83782 by jagoll last updated on 06/Mar/20 $$\mathrm{f}^{\left(\mathrm{5}\right)} \:\left(\mathrm{x}\right)\:=\:\mathrm{4}^{−\mathrm{sin}\:\mathrm{x}} \\ $$$$\mathrm{f}^{\left(\mathrm{7}\right)} \left(\mathrm{x}\right)\:=?\: \\ $$ Commented by john santu last updated on 06/Mar/20 $$\mathrm{f}^{\left(\mathrm{n}\right)}…
Question Number 149252 by holomata last updated on 04/Aug/21 $$ \\ $$$${e}^{{y}} =\left(\mathrm{sin}\:{x}\right)\left(\mathrm{cos}\:{x}\right) \\ $$$${find}\:\frac{{dy}}{{dx}} \\ $$ Answered by Mokmokhi last updated on 04/Aug/21 $$\frac{{dy}}{{dx}}=\mathrm{2cot}\:\mathrm{2}{x}…
Question Number 149191 by mnjuly1970 last updated on 03/Aug/21 $$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:……{calculus}….. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{sech}\left(\pi{x}\right)}{\mathrm{1}+\mathrm{4}{x}^{\:\mathrm{2}} }\:{dx}=? \\ $$$$\:………{m}.{n}… \\ $$ Answered by mindispower last…