Q. 74 - Calculus

Question

$$\begin{gathered} \mathrm{Let} p\left(x\right) \mathrm{be} \mathrm{a} \mathrm{polynomial} \mathrm{of} \mathrm{degree} n, q\left(y\right) \mathrm{be} \mathrm{any} \mathrm{function} \mathrm{of} y, \mathrm{and} f(x, y) = p\left(x\right)q\left(y\right). \\ \mathrm{Prove} \mathrm{that} \frac{{\partial }^{\mathrm{n}+1}\mathrm{f}}{\partial {\mathrm{x}}^{\mathrm{n}+1}} = 0 \end{gathered}$$

Step-by-Step Solution

Answer

$\frac{{\partial }^{\mathrm{n}+1}\mathrm{f}}{\partial {\mathrm{x}}^{\mathrm{n}+1}} = 0$

Step 1. Given information

$$\begin{gathered} f(x, y) = p\left(x\right)q\left(y\right), \\ \mathrm{where}, p\left(x\right) \mathrm{is} \mathrm{a} \mathrm{polynomial} \mathrm{of} \mathrm{degree} n, q\left(y\right) \mathrm{is} \mathrm{any} \mathrm{function} \mathrm{of} y. \end{gathered}$$

Step 2. Proofs of given partial derivatives

$$\begin{gathered} \mathrm{LHS}=\frac{{\partial }^{\mathrm{n}+1}\mathrm{f}}{\partial {\mathrm{x}}^{\mathrm{n}+1}} \\ =\frac{\partial }{\partial \mathrm{x}}\left(\frac{{\partial }^{\mathrm{n}} \mathrm{p}\left(\mathrm{x}\right)\mathrm{q}\left(\mathrm{y}\right)}{\partial {\mathrm{x}}^{\mathrm{n}}}\right) \\ \mathrm{Using} \mathrm{definition} \mathrm{of} \mathrm{partial} \mathrm{derivatives}, \mathrm{function} \mathrm{of} \mathrm{y} \mathrm{is} \mathrm{treated} \mathrm{as} \mathrm{a} \mathrm{constant} \mathrm{term} \\ =\mathrm{q}\left(\mathrm{y}\right)\frac{\partial }{\partial \mathrm{x}}\left(\frac{{\partial }^{\mathrm{n}} \mathrm{p}\left(\mathrm{x}\right)}{\partial {\mathrm{x}}^{\mathrm{n}}}\right) \\ \mathrm{Using} \mathrm{power} \mathrm{rule}, \mathrm{after} \mathrm{differentiating} \mathrm{polynomial} \mathrm{of} \mathrm{degree} n \mathrm{for} n \mathrm{times} \mathrm{we} \mathrm{get} \\ \mathrm{a} \mathrm{constant} \mathrm{term}, \mathrm{whose} \mathrm{differentiation} \mathrm{gives} \mathrm{zero} \mathrm{as} \mathrm{the} \mathrm{final} \mathrm{result} \mathrm{after} \left(\mathrm{n}+1\right) \\ \mathrm{times} \mathrm{of} \mathrm{differentiation} \mathrm{in} \mathrm{total}. \\ = \mathrm{q}\left(\mathrm{y}\right)\frac{\partial \mathrm{K}}{\partial \mathrm{x}} \\ = \mathrm{q}\left(\mathrm{y}\right) \left(0\right) \\ =0 \\ =\mathrm{RHS} \end{gathered}$$