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Question Number 184040 by pete last updated on 02/Jan/23

$$\mathrm{Use}\mathrm{implicit}\mathrm{differentiation}\mathrm{to}\mathrm{find}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}$$$$\mathrm{for}\sin y=x$$

Answered by cortano2 last updated on 02/Jan/23

$$y^{\prime }\cos y=1$$$$y^{″}(-\sin y)=0$$$$y^{″}=0$$

Answered by Yhusuph last updated on 02/Jan/23

$$$$$$\frac{dy}{dx}cosy=1$$$$$$$$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$$$$$$$$\frac{d^{2}y}{{dx}^2}=\frac{-(-2{(1-x^2)}^{-\frac{1}{2}})}{1-x^2}$$$$$$$$\frac{d^{2}y}{{dx}^2}=\frac{2}{{(1-x^2)}^{\frac{3}{2}}}$$$$\frac{{\mathit{d}}^2\mathit{y}}{{\mathit{d}\mathit{x}}^2}=\frac{2}{{\left({}^2\sqrt{1-{\mathit{x}}^2}\right)}^3}$$$$\partial enkenLast\beta orn$$$$Mentor:Proffyemphy$$$$$$

Commented by Frix last updated on 02/Jan/23

$$\mathrm{This}\mathrm{cannot}\mathrm{be}\mathrm{true}.$$$$y=\arcsin x\Rightarrow \frac{d^{2}y}{{dx}^2}=\frac{x}{{(1-x^2)}^{\frac{3}{2}}}$$

Answered by mr W last updated on 03/Jan/23

$$\sin y=x$$$$y^{\prime }\cos y=1$$$$y^{″}\cos y-{\left(y^{\prime }\right)}^2\sin y=0$$$$y^{″}={\left(y^{\prime }\right)}^2\frac{\sin y}{\cos y}=\frac{\tan y}{{\cos }^{2}y}=\frac{\sin y}{{\cos }^{3}y}=\frac{\sin y}{{(1-{\sin }^{2}y)}^{\frac{3}{2}}}$$$$y^{″}=\frac{x}{{(1-x^2)}^{\frac{3}{2}}}$$

Commented by pete last updated on 02/Jan/23

$$\mathrm{Thank}\mathrm{you}\mathrm{Mr}.\mathrm{W},\mathrm{but}\mathrm{the}\mathrm{answer}\mathrm{given}$$$$\mathrm{is}:\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}={\sec }^2\mathrm{ytany}$$

Commented by cortano1 last updated on 03/Jan/23

$$from\sin y=xweget\{\begin{array}{l}\sec y=\frac{1}{\sqrt{1-x^2}}\\ \tan y=\frac{x}{\sqrt{1-x^2}}\end{array}$$$$now{y}^{″}=\frac{x}{(1-x^2)\sqrt{1-x^2}}$$$$={\left[\frac{1}{\sqrt{1-x^2}}\right]}^2.\frac{x}{\sqrt{1-x^2}}=\sec {}^{2}y\tan y$$

Commented by mr W last updated on 03/Jan/23

$$youcanexpress{y}^{″}intermsofyor$$$$intermsofx.youseebothabove.$$$$y^{″}={\left(y^{\prime }\right)}^2\frac{\sin y}{\cos y}=\frac{\tan y}{{\cos }^{2}y}=\frac{\sin y}{{\cos }^{3}y}=\frac{\sin y}{{(1-{\sin }^{2}y)}^{\frac{3}{2}}}$$$$y^{″}=\frac{x}{{(1-x^2)}^{\frac{3}{2}}}$$

Commented by pete last updated on 04/Jan/23

$$\mathrm{Thanks}\mathrm{very}\mathrm{much}\mathrm{sir}$$

Answered by manxsol last updated on 02/Jan/23

Commented by mr W last updated on 03/Jan/23

$$maybenotallpeoplecanreadyour$$$$solutionproperly.thisishowyour$$$$postlookslike:$$

Commented by mr W last updated on 03/Jan/23

Commented by ARUNG_Brandon_MBU last updated on 03/Jan/23

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