Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises - Page 143: 102

Answer

a) See proof b) $ F'''=^3C_0 f'''g+^3C_1f''g'+^3C_2f'g''+^3C_3fg'''$ $F^{iv}=^4C_0f^{iv}g+^4C_1f^{iii}g^{1}+^4C_2f^{ii}g^{ii}+^4C_3f^{1}g^{iii}+^4C_4fg^{iv}$ c) $F^n =^nC_0 f^ng+^nC_1f^{n-1}g^1+^nC_2f^{n-2}g^2+....^nC_rf^{n-r}g^r+...+^nC_nfg^n$

Work Step by Step

$$F(x)=f(x)g(x)$$ Using product rule of differentiation $$F'(x)=f'(x)g(x)+g'(x)f(x)$$ Again using product rule $$F''(x)=\frac{d[f'(x)g(x)]}{dx}+\frac{d[f(x)g'(x)]}{dx}=[f''(x)g(x)+f'(x)g'(x)]+[f'(x)g'(x)+f(x)g''(x)]$$ $$F''(x)=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x)$$ (b)Since $F''=f'''g+3f''g+3f'g''+fg'''$ $$ F'''=^3C_0 f'''g+^3C_1f''g'+^3C_2f'g''+^3C_3fg'''$$ $$F^{iv}=^4C_0f^{iv}g+^4C_1f^{iii}g^{1}+^4C_2f^{ii}g^{ii}+^4C_3f^{1}g^{iii}+^4C_4fg^{iv}$$ Similarly $$F^n =^nC_0 f^ng+^nC_1f^{n-1}g^1+^nC_2f^{n-2}g^2+....+^nC_rf^{n-r}g^r+...+^nC_nfg^n$$
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