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When-y-ax-b-is-a-tangent-line-to-the-curve-f-x-x-3-passing-through-0-2-find-a-b-

Question Number 103659 by abony1303 last updated on 16/Jul/20 $$\mathrm{When}\:\mathrm{y}=\mathrm{ax}+\mathrm{b}\:\mathrm{is}\:\mathrm{a}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{curve}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} \:\mathrm{passing}\:\mathrm{through}\:\left(\mathrm{0};\:−\mathrm{2}\right), \\ $$$$\mathrm{find}\:\mathrm{a}+\mathrm{b}? \\ $$ Commented by abony1303 last updated on 16/Jul/20 $$\mathrm{pls}\:\mathrm{help}…

Question-169164

Question Number 169164 by 0731619 last updated on 25/Apr/22 Commented by infinityaction last updated on 25/Apr/22 $${y}\:=\:\mathrm{ln}\:\left(\sqrt{\frac{\left(\mathrm{1}−\mathrm{sin}\:{x}\right)/\mathrm{cos}\:{x}}{\left(\mathrm{1}+\mathrm{sin}\:{x}\right)/\mathrm{cos}\:{x}}}\right. \\ $$$${y}\:=\:\mathrm{ln}\:\sqrt{\frac{\left(\mathrm{sec}\:{x}−\mathrm{tan}\:{x}\right)\left(\mathrm{sec}\:{x}−\mathrm{tan}\:{x}\right)}{\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)\left(\mathrm{sec}\:{x}−\mathrm{tan}\:{x}\right)}} \\ $$$${y}\:=\:\mathrm{ln}\:\left(\mathrm{sec}\:{x}−\mathrm{tan}\:{x}\right) \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{sec}\:{x}\left(\mathrm{tan}\:{x}−\mathrm{sec}\:{x}\right)}{\left(\mathrm{sec}\:{x}−\mathrm{tan}\:{x}\right)} \\ $$$$\frac{{dy}}{{dx}}\:=\:−\mathrm{sec}\:{x}…

ABC-is-a-triangle-in-which-the-bisector-of-angle-at-B-meet-the-side-AC-at-D-and-the-bisector-of-the-angle-BDC-is-parallel-to-the-side-AB-Prove-that-the-ABC-is-issoceles-triangle-

Question Number 169145 by MathsFan last updated on 24/Apr/22 $$\boldsymbol{{ABC}}\:\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{triangle}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{bisector}} \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{angle}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{{B}}\:\boldsymbol{\mathrm{meet}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{side}}\:\boldsymbol{{AC}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{{D}}, \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{bisector}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{angle}}\:\boldsymbol{{BDC}}\:\boldsymbol{\mathrm{is}} \\ $$$$\:\boldsymbol{\mathrm{parallel}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{side}}\:\boldsymbol{{AB}}.\:\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{that}} \\ $$$$\:\boldsymbol{\mathrm{the}}\:\bigtriangleup\boldsymbol{{ABC}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{issoceles}}\:\boldsymbol{\mathrm{triangle}}. \\ $$$$ \\ $$ Answered by som(math1967)…

Question-169112

Question Number 169112 by mokys last updated on 24/Apr/22 Commented by aleks041103 last updated on 24/Apr/22 $${y}=\left(\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{3}} −\mathrm{4}{x}}\right)^{\mathrm{3}} \\ $$$${y}'=\mathrm{3}\left(\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{3}} −\mathrm{4}{x}}\right)^{\mathrm{2}} \frac{\mathrm{2}{x}\left({x}^{\mathrm{3}} −\mathrm{4}{x}\right)−\left(\mathrm{3}{x}^{\mathrm{2}}…