(a) Check whether the function f ( x )= x^2| x | is differentiable at x =0 or not.

$$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c} \mathrm{x}^{3}, \mathrm{x} \geq 0 \\ -\mathrm{x}^{3}, \mathrm{x} \leq 0 \end{array}\right. $$ $\displaystyle \because R H D=L H D=0, \quad$ So $\displaystyle \mathrm{f}(\mathrm{x})$ is differentiable at $\displaystyle \mathrm{x}=0$ $$\begin{aligned} & \mathrm{y}=\sqrt{\tan \sqrt{\mathrm{x}}} \\ & \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{\sec ^{2} \sqrt{\mathrm{x}}}{2 \sqrt{\tan \sqrt{\mathrm{x}}}} \times \frac{1}{2 \sqrt{\mathrm{x}}} \\ & \begin{aligned} \sqrt{\mathrm{x}} \frac{d \mathrm{y}}{d \mathrm{x}} & =\frac{\sec ^{2} \sqrt{\mathrm{x}}}{4 \sqrt{\tan \sqrt{\mathrm{x}}}} \\ & =\frac{1+(\tan \sqrt{\mathrm{x}})^{2}}{4 \sqrt{\tan \sqrt{\mathrm{x}}}}=\frac{1+\mathrm{y}^{4}}{4 \mathrm{y}} \end{aligned} \end{aligned} $$