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Question

# If y log (1 + cos x), prove that $\frac{{d}^{3}y}{d{x}^{3}}+\frac{{d}^{2}y}{d{x}^{2}}·\frac{dy}{dx}=0$

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Solution

## Here, $y=\mathrm{log}\left(1+\mathrm{cos}x\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\frac{\mathit{d}y}{\mathit{d}x}=\frac{-\mathrm{sin}x}{1+\mathrm{cos}x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\frac{{\mathit{d}}^{2}y}{\mathit{d}{x}^{2}}=\frac{-\mathrm{cos}x-{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}{{\left(1+\mathrm{cos}x\right)}^{2}}=\frac{-\left(\mathrm{cos}x+1\right)}{\left(1+\mathrm{cos}x\right)}=\frac{-1}{1+\mathrm{cos}x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\frac{{\mathit{d}}^{3}y}{\mathit{d}{x}^{3}}=\frac{-\mathrm{sin}x}{{\left(1+\mathrm{cos}x\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{{\mathit{d}}^{3}y}{\mathit{d}{x}^{3}}+\frac{\mathrm{sin}x}{{\left(1+\mathrm{cos}x\right)}^{2}}=0\phantom{\rule{0ex}{0ex}}⇒\frac{{\mathit{d}}^{3}y}{\mathit{d}{x}^{3}}+\left(\frac{-1}{1+\mathrm{cos}x}\right)\left(\frac{-\mathrm{sin}x}{1+\mathrm{cos}x}\right)=0\phantom{\rule{0ex}{0ex}}⇒\frac{{\mathit{d}}^{3}y}{\mathit{d}{x}^{3}}+\frac{{\mathit{d}}^{2}y}{\mathit{d}{x}^{2}}×\frac{\mathit{d}y}{\mathit{d}x}=0$

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