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Question Number 133334 by mnjuly1970 last updated on 21/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:\:\:\:\:{calculus}… \\ $$$$\:{prove}\:\:{that}:: \\ $$$$\:\:\:\boldsymbol{\phi}=\int_{−\infty} ^{\:+\infty} \frac{\:{cosh}\left({px}\right)}{{cosh}\left({x}\right)}\:=\:\frac{\pi}{{cos}\left(\frac{\pi{p}}{\mathrm{2}}\right)} \\ $$$$ \\ $$ Answered by mnjuly1970 last updated…
Question Number 2229 by tabrez8590@gmail last updated on 09/Nov/15 $${f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{10} \\ $$$$\Rightarrow{f}^{'} \left({x}\right)=\mathrm{2}{x}+\mathrm{3}\:{and} \\ $$$${f}^{'} \left(\mathrm{1}\right)=\mathrm{5}\:{what}\:{indicate}\:{these}\:{vslue}\: \\ $$$${f}'\left({x}\right){andf}'\left(\mathrm{1}\right)\:{in}\:{geometrical}\:{meaning} \\ $$ Answered by prakash jain…
Question Number 67760 by ugwu Kingsley last updated on 31/Aug/19 $${y}'={y}^{\mathrm{2}} +\mathrm{2}\:;\:{y}\left(\mathrm{0}\right)=\mathrm{2} \\ $$$$ \\ $$$${solve}\:{by}\:{picards}\:{iteration}\:{method} \\ $$$$ \\ $$ Terms of Service Privacy Policy…
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Question Number 2219 by tabrez8590@gmail last updated on 09/Nov/15 $${it}\:{is}\:{necessary}\:{for}\:{diffrentiable}\:\:{function}\:{itmust}\:{becontinuty}\:{why}?{and}\:{how}\:{please}\:{explain}\:{by}\:{example} \\ $$$$ \\ $$ Answered by Filup last updated on 09/Nov/15 $$\mathrm{if}\:\begin{cases}{{y}={x}\:\:\:\mathrm{if}\:{x}>\mathrm{0}}\\{{y}=−{x}\:\:\:\mathrm{if}\:{x}<\mathrm{0}}\end{cases} \\ $$$$ \\…
Question Number 133194 by Algoritm last updated on 19/Feb/21 Commented by Algoritm last updated on 19/Feb/21 $$\frac{\mathrm{d}^{\mathrm{n}} \mathrm{y}}{\mathrm{dx}^{\mathrm{n}} }=? \\ $$ Commented by mr W…
Question Number 67653 by mhmd last updated on 29/Aug/19 Answered by mr W last updated on 29/Aug/19 $${let}\:{t}=\frac{{y}}{{x}} \\ $$$${y}={xt} \\ $$$${y}'={t}+{xt}' \\ $$$$ \\…
Question Number 133181 by mnjuly1970 last updated on 19/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……{nice}\:\:\:\:\:{calculus}… \\ $$$$\:\:\:\:\:\:{lim}\:_{{n}\rightarrow\infty} \left\{{n}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{{n}+{k}}\right)^{\mathrm{2}} \right\}=?? \\ $$ Answered by Dwaipayan Shikari last updated on…
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