Chapter 12: Q. 74 (page 946)

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

Short Answer

Expert verified

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

Step by step solution

Step 1. Given information

$f (x, y) = p (x) q (y), where, p (x) is a polynomial of degree n, q (y) is any function of y .$

Step 2. Proofs of given partial derivatives

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

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

Short Answer

Step by step solution

Step 1. Given information

Step 2. Proofs of given partial derivatives

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Letp(x)beapolynomialofdegreen,q(y)beanyfunctionofy,andf(x,y)=p(x)q(y).Provethat∂n+1f∂xn+1=0

Short Answer

Step by step solution

Step 1. Given information

Step 2. Proofs of given partial derivatives

One App. One Place for Learning.

Most popular questions from this chapter

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