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Q. 88

Expert-verified

Calculus

Found in: Page 199

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Use the definition of the derivative to prove the following special case of the product rule
$\frac{d}{d x} (x^{2} f (x)) = 2 x f (x) + x^{2} f' (x)$

We proved the special case of product function using the definition of the derivative

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information

We are given a function $\frac{d}{d x} (x^{2} f (x)) = 2 x f (x) + x^{2} f' (x)$

Step 2: Find the derivative

Consider $g (x) = x^{2} f (x)$

Using the definition of derivative we get,

$\lim_{h \to 0} \frac{g (x + h) - g (x)}{h} \lim_{h \to 0} \frac{{(x + h)}^{2} f (x + h) - x^{2} f (x)}{h} \lim_{h \to 0} \frac{(x^{2} + 2 x h + h^{2}) f (x + h) - x^{2} f (x)}{h} \lim_{h \to 0} \frac{x^{2} (f (x + h) - f (x))}{h} + \lim_{h \to 0} \frac{h (2 x + h) f (x + h)}{h} = x^{2} f' (x) + 2 x f (x)$

Hence proved

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