Question Number 171529 by cortano1 last updated on 17/Jun/22 Commented by kapoorshah last updated on 17/Jun/22 $$\mathrm{25} \\ $$ Answered by FelipeLz last updated on…
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Question Number 40319 by ajfour last updated on 19/Jul/18 $${y}=\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{{a}} \\ $$$${Find}\:\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\:. \\ $$ Commented by math khazana by abdo…
Question Number 171379 by mathlove last updated on 14/Jun/22 $${f}\left({x}\right)=\begin{cases}{\mathrm{2}{x}+\mathrm{3}\:\:\:;{x}>\mathrm{0}}\\{\mathrm{3}{x}−\mathrm{5}\:\:;{x}\leqslant\mathrm{0}}\end{cases} \\ $$$$\frac{{df}\left({x}\right)}{{dx}}=? \\ $$ Commented by MJS_new last updated on 14/Jun/22 $${f}\left({x}\right)\:=\begin{cases}{\mathrm{3}{x}−\mathrm{5};\:{x}\leqslant\mathrm{0}}\\{\mathrm{not}\:\mathrm{defined};\:\mathrm{0}<{x}\leqslant\mathrm{2}}\\{\mathrm{2}{x}+\mathrm{3};\:{x}>\mathrm{2}}\end{cases} \\ $$$${f}\:'\left({x}\right)=\begin{cases}{\mathrm{3};\:{x}\leqslant\mathrm{0}}\\{\mathrm{not}\:\mathrm{defined};\:\mathrm{0}<{x}\leqslant\mathrm{2}}\\{\mathrm{2};\:{x}>\mathrm{2}}\end{cases} \\…
Question Number 40277 by Raj Singh last updated on 18/Jul/18 Answered by MJS last updated on 18/Jul/18 $$\frac{{d}}{{d}\theta}\left[\mathrm{sin}^{{p}} \:\theta\:\mathrm{cos}^{{q}} \:\theta\right]=\mathrm{sin}^{{p}−\mathrm{1}} \:\theta\:\mathrm{cos}^{{q}−\mathrm{1}} \:\theta\:\left({p}\mathrm{cos}^{\mathrm{2}} \:\theta\:−{q}\mathrm{sin}^{\mathrm{2}} \:\theta\right) \\…
Question Number 40276 by Raj Singh last updated on 18/Jul/18 Answered by MJS last updated on 18/Jul/18 $$\frac{{d}}{{dx}}\left[\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{tan}\:{x}}\right]=\frac{\mathrm{cos}^{\mathrm{3}} \:{x}\:−\mathrm{sin}^{\mathrm{3}} \:{x}}{\mathrm{1}+\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}} \\ $$$$\mathrm{zeros}\:\mathrm{at}\:{x}=\frac{\pi}{\mathrm{4}}\left(\mathrm{4}{k}+\mathrm{1}\right)\:\wedge\:{k}\in\mathbb{Z} \\ $$$$\mathrm{max}\:\mathrm{at}\:{x}=\frac{\pi}{\mathrm{4}}\left(\mathrm{4}{k}+\mathrm{1}\right)\:\wedge\:{k}=\mathrm{2}{n}\:\wedge\:{n}\in\mathbb{Z}\:\Rightarrow \\…
Question Number 171315 by pete last updated on 12/Jun/22 $$\mathrm{The}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{y}=\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{2} \\ $$$$\mathrm{at}\:\left(\mathrm{1},\frac{\mathrm{1}}{\mathrm{2}}\right)\:\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{normal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\mathrm{y}=\mathrm{x}^{\mathrm{2}} +\mathrm{6x}+\mathrm{10}\:\mathrm{at}\:\left(−\mathrm{2},\mathrm{2}\right).\:\mathrm{Find}\:\mathrm{the}\:\mathrm{values} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}. \\ $$ Answered by som(math1967) last updated…
Question Number 171267 by pete last updated on 11/Jun/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{coordinates}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\mathrm{y}=\frac{{x}}{\mathrm{1}+{x}}\:\mathrm{at}\:\mathrm{which}\:\mathrm{the}\:\mathrm{tangents}\:\mathrm{to}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\mathrm{are}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{line}\:{x}−{y}+\mathrm{8}=\mathrm{0}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tangents}\:\mathrm{at} \\ $$$$\mathrm{these}\:\mathrm{points}. \\ $$ Commented by greougoury555 last updated…
Question Number 171265 by ali009 last updated on 11/Jun/22 $${find}\:{the}\:{drivative}\:{of}\: \\ $$$${f}\left({x},{y},{z}\right)={cos}\left({xy}\right)+{e}^{{zy}} +{ln}\left({zy}\right) \\ $$$${at}\:{point}\:\left(\mathrm{1},\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right)\:{in}\:{the}\:{direction} \\ $$$${v}={i}+\mathrm{2}{j}+\mathrm{2}{k} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 105706 by bramlex last updated on 31/Jul/20 $$\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{{dx}}{\left(\sqrt{\mathrm{5}}−\mathrm{cos}\:{x}\right)^{\mathrm{3}} }\:? \\ $$ Answered by john santu last updated on 31/Jul/20 $${F}\left({a}\right)\:=\:\underset{\mathrm{0}} {\overset{\pi}…