Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 72

Answer

$$f''(x)=6xg'(x^2)+4x^3g''(x^2)$$

Work Step by Step

$$f(x)=xg(x^2)$$ 1) First derivative of $f(x)$ $$f'(x)=x'g(x^2)+x[g(x^2)]'$$ $$f'(x)=g(x^2)+x[g(x^2)]'$$ - Considering $[g(x^2)]'$ $$[g(x^2)]'=\frac{dg(x^2)}{dx}$$ Apply the Chain Rule: $$[g(x^2)]'=\frac{dg(x^2)}{d(x^2)}\frac{d(x^2)}{dx}$$ $$[g(x^2)]'=g'(x^2)\times2x$$ $$[g(x^2)]'=2xg'(x^2)$$ Therefore, $$f'(x)=g(x^2)+2x^2g'(x^2)$$ 2) Second derivative of $f(x)$ $$f''(x)=[g(x^2)+2x^2g'(x^2)]'$$ $$f''(x)=[g(x^2)]'+2[(x^2)'g'(x^2)+x^2[g'(x^2)]']$$ - Considering $[g'(x^2)]'$ $$[g'(x^2)]'=\frac{dg'(x^2)}{dx}=\frac{dg'(x^2)}{d(x^2)}\frac{d(x^2)}{dx}$$ (Chain Rule) $$[g'(x^2)]'=2xg''(x^2)$$ - Also, from part 1), we prove that $[g(x^2)]'=2xg'(x^2)$ - Also, $(x^2)'=2x$ Therefore, $$f''(x)=2xg'(x^2)+2[2xg'(x^2)+2x^3g''(x^2)]$$ $$f''(x)=2xg'(x^2)+4xg'(x^2)+4x^3g''(x^2)$$ $$f''(x)=6xg'(x^2)+4x^3g''(x^2)$$
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