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Question Number 96782 by  M±th+et+s last updated on 04/Jun/20

$$1)\frac{{cos}^4\left(\theta \right)}{x}-\frac{{sin}^4\left(\theta \right)}{y}=\frac{1}{x+y}$$$$find\frac{dy}{dx}$$$$$$$$2)solve:2\lfloor x-4+\lfloor x\rfloor \rfloor =6-3\lfloor x\rfloor$$$$$$$$3)\underset{x\to 4}{lim}\frac{{\left(cos\left(x\right)\right)}^x-{\left(sin\left(x\right)\right)}^x-cos\left(2x\right)}{(x-4)}$$$$$$$$$$

Commented by  M±th+et+s last updated on 04/Jun/20

$$thanksforsolutions$$

Answered by mr W last updated on 04/Jun/20

$$\left(2\right)$$$$\lfloor x\rfloor mustbeeven,sayx=2n+f$$$$2\lfloor 2n+f-4+2n\rfloor =6-3\times 2n$$$$\lfloor 4(n-1)+f\rfloor =3(1-n)$$$$7n=7$$$$\Rightarrow n=1$$$$\Rightarrow 2⩽x<3$$

Answered by Sourav mridha last updated on 04/Jun/20

$$\left(3\right)\frac{0}{0}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{s}\mathit{o}\mathit{u}\mathit{s}\mathit{i}\mathit{n}\mathit{g}{\mathit{L}}^{\prime }\mathit{H}\hat{\mathrm{o}}\mathit{p}\mathit{i}\mathit{t}\mathit{a}\mathit{l}$$$$\mathit{A}\mathit{n}\mathit{s}:{\mathit{c}\mathit{o}\mathit{s}}^4\left(4\right)[-4\mathit{t}\mathit{a}\mathit{n}\left(4\right)+\mathit{l}\mathit{n}\left(\mathit{c}\mathit{o}\mathit{s}\left(4\right)\right)]$$$$-{\mathit{s}\mathit{i}\mathit{n}}^4\left(4\right)[4\mathit{c}\mathit{o}\mathit{t}\left(4\right)+\mathit{l}\mathit{n}\left(\mathit{s}\mathit{i}\mathit{n}\left(4\right)\right)]$$$$+2\mathit{s}\mathit{i}\mathit{n}\left(8\right)..$$

Answered by abdomathmax last updated on 05/Jun/20

$$1)\left(\mathrm{e}\right)\Rightarrow \frac{{\cos }^4\theta }{\mathrm{x}}=\frac{{\sin }^4\theta }{\mathrm{y}}+\frac{1}{\mathrm{x}+\mathrm{y}}\Rightarrow$$$$\frac{\mathrm{y}}{\mathrm{x}}{\cos }^4\theta ={\sin }^4\theta +\frac{\mathrm{y}}{\mathrm{x}+\mathrm{y}}\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}{\cos }^4\theta ={\sin }^4\theta +\frac{1}{(\frac{\mathrm{x}}{\mathrm{y}}+1)}$$$$\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{u}\Rightarrow \frac{{\cos }^4\theta }{\mathrm{u}}={\sin }^4\theta +\frac{1}{\mathrm{u}+1}\Rightarrow$$$$\frac{{\cos }^4\theta }{\mathrm{u}}-\frac{1}{\mathrm{u}+1}={\sin }^4\theta \Rightarrow \frac{{\cos }^4\theta \mathrm{u}+{\cos }^4\theta -\mathrm{u}}{\mathrm{u}(\mathrm{u}+1)}={\sin }^4\theta$$$$\Rightarrow {\cos }^4\theta \mathrm{u}+{\cos }^4\theta -\mathrm{u}={\sin }^4\theta {\mathrm{u}}^2+{\sin }^4\theta \mathrm{u}\Rightarrow$$$${\sin }^4\theta {\mathrm{u}}^2+({\sin }^4-{\cos }^4\theta +1)\mathrm{u}-{\cos }^4\theta =0\Rightarrow$$$${\sin }^4\theta {\mathrm{u}}^2+({\sin }^2\theta -{\cos }^2\theta +1)\mathrm{u}-{\cos }^4\theta =0$$$$\mathrm{\Delta }={({\sin }^2\theta -{\cos }^2\theta +1)}^2+4{\sin }^4\theta {\cos }^4\theta$$$$\mathrm{u}=\frac{-({\sin }^2\theta -{\cos }^2\theta +1)\stackrel{-}{+}\sqrt{\mathrm{\Delta }}}{{2\sin }^4\theta }={\mathrm{A}}_{\theta }$$$$\mathrm{y}=\frac{\mathrm{A}\theta }{\mathrm{x}}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{{\mathrm{A}}_{\theta }}{{\mathrm{x}}^2}$$

Commented by abdomathmax last updated on 05/Jun/20

$$\mathrm{sorry}\mathrm{y}=\frac{\mathrm{x}}{\mathrm{u}}=\frac{\mathrm{x}}{{\mathrm{A}}_{\theta }}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{{\mathrm{A}}_{\theta }}$$

Commented by  M±th+et+s last updated on 05/Jun/20

$$welldonesir$$

Commented by mathmax by abdo last updated on 05/Jun/20

$$\mathrm{thankx}$$

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