Question Number 141769 by 7770 last updated on 23/May/21
$$\boldsymbol{\mathrm{If}}\:\:\boldsymbol{\mathrm{y}}=\mathrm{2}\boldsymbol{\mathrm{sinx}}+\boldsymbol{\mathrm{tanx}},\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}} \\ $$$$\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }=\mathrm{2}\boldsymbol{\mathrm{sinx}}\left(\boldsymbol{\mathrm{sec}}^{\mathrm{3}} \boldsymbol{\mathrm{x}}−\mathrm{1}\right) \\ $$$$\boldsymbol{\mathrm{Any}}\:\boldsymbol{\mathrm{detailed}}\:\boldsymbol{\mathrm{solution}}\:\boldsymbol{\mathrm{please}}. \\ $$
Answered by physicstutes last updated on 23/May/21
$${y}\:=\:\mathrm{2}\:\mathrm{sin}\:{x}\:+\:\mathrm{tan}\:{x} \\ $$$$\frac{{dy}}{{dx}}\:=\:\mathrm{2}\:\mathrm{cos}\:{x}\:+\:\mathrm{sec}^{\mathrm{2}} {x} \\ $$$$\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:−\mathrm{2}\:\mathrm{sin}\:{x}\:+\frac{\left(\mathrm{cos}^{\mathrm{2}} {x}\right)\left(\mathrm{0}\right)−\left(\mathrm{1}\right)\left(\mathrm{2cos}\:{x}\right)\left(−\mathrm{sin}\:{x}\right)}{\mathrm{cos}^{\mathrm{4}} {x}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:\frac{\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}{\mathrm{cos}^{\mathrm{4}} {x}}\:−\:\mathrm{2}\:\mathrm{sin}\:{x}\:=\:\frac{\mathrm{2sin}\:{x}}{\mathrm{cos}^{\mathrm{3}} {x}}−\mathrm{2sin}\:{x}\:=\:\mathrm{2sin}\:{x}\left(\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{3}} {x}}−\mathrm{1}\right)\:=\:\mathrm{2sin}\:{x}\left(\mathrm{sec}^{\mathrm{3}} {x}−\mathrm{1}\right) \\ $$$$\mathrm{as}\:\mathrm{required}. \\ $$