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Question Number 45456 by peter frank last updated on 13/Oct/18

Commented by peter frank last updated on 13/Oct/18

$$\mathrm{math}1967$$

Commented by math1967 last updated on 13/Oct/18

$$y=\frac{sinx}{x}$$$$xy=sinx\Rightarrow y+x\frac{dy}{dx}=cosx$$$$\frac{dy}{dx}+\frac{dy}{dx}+x\frac{d^{2}y}{{dx}^2}=-sinx$$$$x\frac{d^{2}y}{{dx}^2}+2\frac{dy}{dx}=-xy[xy=sinx]$$$$x^2\frac{d^{2}y}{{dx}^2}+2x\frac{dy}{dx}+x^{2}y=0$$$$cannotmatch....$$

Commented by maxmathsup by imad last updated on 14/Oct/18

$$2)siny=2sinx\Rightarrow y=arcsin\left(2sinx\right)\Rightarrow \frac{dy}{dx}=\frac{2cosx}{\sqrt{1-4{sin}^{2}x}}\Rightarrow$$$${\left(\frac{dy}{dx}\right)}^2=\frac{4{cos}^{2}x}{1-4{sin}^{2}x}=\frac{4(1-{sin}^{2}x)}{1-4{sin}^{2}x}=\frac{4-4\frac{{sin}^2\left(y\right)}{4}}{1-{sin}^2\left(y\right)}=\frac{4-{sin}^{2}y}{1-{sin}^{2}y}$$$$=\frac{1-{sin}^{2}y+3}{1-{sin}^{2}y}=1+\frac{3}{{cos}^2\left(y\right)}=1+3(1+{tan}^{2}y).$$

Commented by peter frank last updated on 14/Oct/18

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.\mathrm{i}\mathrm{also}\mathrm{had}\mathrm{such}\mathrm{doubt}$$

Commented by peter frank last updated on 14/Oct/18

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18

$$2)cosy\frac{dy}{dx}=2cosx$$$${\left(\frac{dy}{dx}\right)}^2=\frac{4{cos}^{2}x}{{cos}^{2}y}=\frac{4(1-{sin}^{2}x)}{{cos}^{2}y}$$$$=\frac{4(1-\frac{{sin}^{2}y}{4})}{{cos}^{2}y}=\frac{4(4-{sin}^{2}y)}{4{cos}^{2}y}=\frac{4(4-1+{cos}^{2}y)}{4{cos}^{2}y}$$$$=\frac{3{sec}^{2}y+1}{1}$$

Commented by peter frank last updated on 14/Oct/18

$$\mathrm{thank}\mathrm{you}$$

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