Question

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

Answer 1

Here,

$$\begin{gathered} y=\log \left(1+\cos x\right) \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}y}{\mathit{d}x}=\frac{-\sin x}{1+\cos x} \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{2}y}{\mathit{d}{x}^2}=\frac{-\cos x-{\cos }^{2}x-{\sin }^{2}x}{{\left(1+\cos x\right)}^2}=\frac{-\left(\cos x+1\right)}{\left(1+\cos x\right)}=\frac{-1}{1+\cos x} \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{3}y}{\mathit{d}{x}^3}=\frac{-\sin x}{{\left(1+\cos x\right)}^2} \\ \Rightarrow \frac{{\mathit{d}}^{3}y}{\mathit{d}{x}^3}+\frac{\sin x}{{\left(1+\cos x\right)}^2}=0 \\ \Rightarrow \frac{{\mathit{d}}^{3}y}{\mathit{d}{x}^3}+\left(\frac{-1}{1+\cos x}\right)\left(\frac{-\sin x}{1+\cos x}\right)=0 \\ \Rightarrow \frac{{\mathit{d}}^{3}y}{\mathit{d}{x}^3}+\frac{{\mathit{d}}^{2}y}{\mathit{d}{x}^2}\times \frac{\mathit{d}y}{\mathit{d}x}=0 \end{gathered}$$