1. **Problem Statement:** Show that $\frac{\partial A_p}{\partial x^2}$ is not a tensor even if $A_p$ is a covariant tensor of rank one. 2. **Recall the definition of a covariant tensor of rank one:** $A_p$ transforms under a coordinate change $x^i \to x^{i'}$ as $$A_{p'} = \frac{\partial x^q}{\partial x^{p'}} A_q.$$ 3. **Consider the derivative $\frac{\partial A_p}{\partial x^2}$:** This is a partial derivative of the tensor component with respect to the coordinate $x^2$. 4. **Transformation of the derivative:** Under coordinate change, $$\frac{\partial A_{p'}}{\partial x^{2'}} = \frac{\partial}{\partial x^{2'}} \left( \frac{\partial x^q}{\partial x^{p'}} A_q \right) = \frac{\partial^2 x^q}{\partial x^{2'} \partial x^{p'}} A_q + \frac{\partial x^q}{\partial x^{p'}} \frac{\partial A_q}{\partial x^{2'}}.$$ 5. **Rewrite $\frac{\partial A_q}{\partial x^{2'}}$ using chain rule:** $$\frac{\partial A_q}{\partial x^{2'}} = \frac{\partial x^m}{\partial x^{2'}} \frac{\partial A_q}{\partial x^m}.$$ 6. **Substitute back:** $$\frac{\partial A_{p'}}{\partial x^{2'}} = \frac{\partial^2 x^q}{\partial x^{2'} \partial x^{p'}} A_q + \frac{\partial x^q}{\partial x^{p'}} \frac{\partial x^m}{\partial x^{2'}} \frac{\partial A_q}{\partial x^m}.$$ 7. **Interpretation:** The presence of the second derivative term $\frac{\partial^2 x^q}{\partial x^{2'} \partial x^{p'}} A_q$ means that $\frac{\partial A_p}{\partial x^2}$ does not transform like a tensor because tensor transformation laws involve only first derivatives of the coordinate transformation. 8. **Conclusion:** Since the transformation law for $\frac{\partial A_p}{\partial x^2}$ includes second derivatives of the coordinate transformation, it is not a tensor, even though $A_p$ itself is a covariant tensor of rank one. This shows explicitly why the partial derivative of a tensor component is not generally a tensor.