(a)OR(b)

If $\displaystyle \mathrm{y}=\sqrt{\cos \mathrm{x}+\mathrm{y}}$, prove that $\displaystyle \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{\sin \mathrm{x}}{1-2 \mathrm{y}}$.

$$\begin{aligned} & \mathrm{y}^{2}=\cos \mathrm{x}+\mathrm{y} & (2 \mathrm{y}-1) \frac{d \mathrm{y}}{d \mathrm{x}}=-\sin \mathrm{x} & \Rightarrow \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{\sin \mathrm{x}}{1-2 \mathrm{y}} \end{aligned} \text { (b) } \mathrm{f}(\mathrm{x})=\left\{\begin{aligned} \mathrm{x}^{3}, & \mathrm{x} \geq 0 -\mathrm{x}^{3}, & \mathrm{x} \leq 0 \end{aligned}\right. \begin{aligned} & \text { At } \mathrm{x}=0 & \text { LHD }=\lim _{h \rightarrow 0} \frac{\mathrm{f}(0-h)-\mathrm{f}(0)}{-h}=\lim _{h \rightarrow 0}\left(\frac{h^{3}}{-h}\right)=\lim _{h \rightarrow 0}\left(-h^{2}\right)=0 \end{aligned} \text { RHD }=\lim _{h \rightarrow 0} \frac{\mathrm{f}(0+h)-\mathrm{f}(0)}{h}=\lim _{h \rightarrow 0}\left(\frac{h^{3}}{h}\right)=\lim _{h \rightarrow 0}\left(h^{2}\right)=0$\displaystyle \because$ LHD $\displaystyle =$ RHD at $\displaystyle \mathrm{x}=0$; when $\displaystyle \mathrm{x} \neq 0, \mathrm{f}(\mathrm{x})$ is a polynomial and hence differentiable. $\displaystyle \therefore \mathrm{f}(\mathrm{x})$ is differentiable at all points.