Question Number 7 by user1 last updated on 25/Jan/15 $$\mathrm{If}\:\:\:{A}=\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{4}}&{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}{and}\:\:{B}=\begin{bmatrix}{\mathrm{1}}&{−\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}}\\{\mathrm{5}}&{\mathrm{0}}\end{bmatrix}, \\ $$$$\mathrm{verify}\:\mathrm{that}\:\:\left(\mathrm{A}{B}\right)'={B}'{A}' \\ $$ Answered by user1 last updated on 30/Oct/14 $$\mathrm{We}\:\mathrm{have}\: \\ $$$${AB}=\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{4}}&{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}\centerdot\begin{bmatrix}{\mathrm{1}}&{−\mathrm{1}}\\{\mathrm{0}}&{\:\:\:\:\mathrm{2}}\\{\mathrm{5}}&{\:\:\:\:\mathrm{0}}\end{bmatrix} \\…
Question Number 131079 by mathlove last updated on 01/Feb/21 $$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\mathrm{2}\sqrt{\mathrm{2}}{x}} {e}^{\mathrm{2}\sqrt{\mathrm{2}}{h}} −{e}^{\mathrm{2}\sqrt{\mathrm{2}}{x}} }{{h}}=? \\ $$ Commented by EDWIN88 last updated on 01/Feb/21 $$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{\mathrm{2}\sqrt{\mathrm{2}}\:\left({x}+{h}\right)}…
Question Number 131078 by Dwaipayan Shikari last updated on 01/Feb/21 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{8}} +\mathrm{1}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 65541 by Tanmay chaudhury last updated on 31/Jul/19 $${we}\:{are}\:{all}\:{students}\:{learning}\:{mutually}\:{from}\:{each}\:{other} \\ $$$${we}\:{are}\:{not}\:{at}\:{par}\:{with}\:{great}\:{Ramanuj}\:{and} \\ $$$${Hardy}…{to}\:{solve}\:{problems}\:{related}\:{to}\:{zeta}\:{function} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 4 by user1 last updated on 25/Jan/15 $$\mathrm{If} \\ $$$$\mathrm{F}\left({y}\frac{\partial{f}}{\partial{z}}−{z}\frac{\partial{f}}{\partial{y}}\right)\boldsymbol{\mathrm{i}}+\left({z}\frac{\partial{f}}{\partial{x}}−{x}\frac{\partial{f}}{\partial{z}}\right)\boldsymbol{\mathrm{j}}+\left({x}\frac{\partial{f}}{\partial{y}}−{y}\frac{\partial{f}}{\partial{x}}\right)\boldsymbol{\mathrm{k}} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\:\:\boldsymbol{\mathrm{F}}=\boldsymbol{\mathrm{r}}×\bigtriangledown{f}. \\ $$ Answered by user1 last updated on 29/Oct/14 $$\:\:\boldsymbol{\mathrm{r}}×\bigtriangledown\boldsymbol{\mathrm{f}}=\begin{vmatrix}{\boldsymbol{\mathrm{i}}}&{\boldsymbol{\mathrm{j}}}&{\boldsymbol{\mathrm{k}}}\\{{x}}&{{y}}&{{z}}\\{\frac{\partial{f}}{\partial{x}}}&{\frac{\partial{f}}{\partial{y}}}&{\frac{\partial{f}}{\partial{z}}}\end{vmatrix} \\…
Question Number 131073 by Algoritm last updated on 01/Feb/21 Answered by mr W last updated on 01/Feb/21 $${F}\:'\left({x}\right)={e}^{\left(\mathrm{ln}\:{x}\right)^{\mathrm{100}} } \frac{\mathrm{1}}{{x}} \\ $$$${F}\:'\left({e}\right)={e}^{\left(\mathrm{ln}\:{e}\right)^{\mathrm{100}} } \frac{\mathrm{1}}{{e}}={e}×\frac{\mathrm{1}}{{e}}=\mathrm{1} \\…
Question Number 65539 by Masumsiddiqui399@gmail.com last updated on 31/Jul/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 2 by user1 last updated on 25/Jan/15 $$\mathrm{Find}\:\mathrm{grad}\:{log}\:\mid\boldsymbol{\mathrm{r}}\mid. \\ $$ Answered by user1 last updated on 29/Oct/14 $$\mathrm{We}\:\mathrm{have}\:\:\:{r}=\sqrt{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)} \\ $$$$\mathrm{log}\:\mid\boldsymbol{\mathrm{r}}\mid=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left({x}^{\mathrm{2}}…
Question Number 131075 by benjo_mathlover last updated on 01/Feb/21 $$\mathrm{If}\:\mathrm{a}−\mathrm{3}=−\mathrm{b}−\mathrm{4}=−\mathrm{c}−\mathrm{5}=\mathrm{d}+\mathrm{6}=\mathrm{e}+\mathrm{7}= \\ $$$$\mathrm{a}−\mathrm{b}−\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{8}\:\mathrm{then}\:\mathrm{a}−\mathrm{b}−\mathrm{c}+\mathrm{d}+\mathrm{e}\:=? \\ $$ Answered by EDWIN88 last updated on 01/Feb/21 $${a}\:={k}+\mathrm{3}\:;\:{b}=−\mathrm{4}−{k}\:;\:{c}=−{k}−\mathrm{5} \\ $$$${d}={k}−\mathrm{6}\:;\:{e}={k}−\mathrm{7} \\…
Question Number 78549 by aliesam last updated on 18/Jan/20 Answered by ~blr237~ last updated on 18/Jan/20 $$\mathrm{let}\:\mathrm{named}\:\mathrm{it}\:\mathrm{A} \\ $$$$\mathrm{state}\:\mathrm{u}=\frac{\mathrm{x}}{\mathrm{2}}\:\:,\:\mathrm{A}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\sqrt{\mathrm{tan2u}}}{\mathrm{1}+\mathrm{sinu}}\:\mathrm{du} \\ $$$$\frac{\mathrm{A}}{\mathrm{2}}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\sqrt{\mathrm{tan2u}}}{\mathrm{cos}^{\mathrm{2}}…