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Question Number 196277 by cortano12 last updated on 21/Aug/23

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Answered by mr W last updated on 21/Aug/23

$$sayz={\sin }^{3}x$$$$\frac{dz}{dy}=\frac{dz}{dx}\times \frac{dx}{dy}=3{\sin }^{2}x\cos x\times \frac{1}{-2\sin 2x}=-\frac{3}{4}\sin x$$$$\frac{d^{2}z}{{dy}^2}=\frac{d}{dy}\left(\frac{dz}{dy}\right)=\frac{d}{dx}\left(\frac{dz}{dy}\right)\times \frac{dx}{dy}=-\frac{3\cos x}{4}\times \frac{1}{-2\sin 2x}$$$$=\frac{3}{16\sin x}✓$$

Answered by deleteduser1 last updated on 21/Aug/23

$$y=1-2{sin}^{2}x\Rightarrow {sin}^{3}x={\left(\frac{1-y}{2}\right)}^{\frac{3}{2}}=\frac{{(1-y)}^{\frac{3}{2}}}{2\sqrt{2}}$$$$\Rightarrow y^{\prime }\left({sin}^{3}x\right)=\frac{-3}{4\sqrt{2}}{(1-y)}^{\frac{1}{2}}\Rightarrow y^{″}\left({sin}^{3}x\right)=\frac{3}{8\sqrt{2}\sqrt{1-y}}$$$$=\frac{3}{8\sqrt{2}\sqrt{2{sin}^{2}x}}=\frac{3}{16sinx}$$

Answered by horsebrand11 last updated on 22/Aug/23

$$\underset{―}{\underbrace{}}$$

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