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 - Page 205: 70


(a) $$g'(0)=c+5$$ $$g''(0)=c^2-2$$ (b) The equation of the tangent line is $$(l): y=(3k+5)x+3$$

Work Step by Step

(a) $$g(x)=e^{cx}+f(x)$$ *The derivative of $g(x)$ is $$g'(x)=(e^{cx})'+f'(x)$$ According to the Chain Rule, $$(e^{cx})'=\frac{d(e^{cx})}{dx}=\frac{d(e^{cx})}{d(cx)}\frac{d(cx)}{dx}$$ $$(e^{cx})'=e^{cx}\times c\times\frac{dx}{dx}$$ $$(e^{cx})'=ce^{cx}$$ Therefore, $$g'(x)=ce^{cx}+f'(x)$$ *Next, we find $g''(x)$ $$g''(x)=(ce^{cx}+f'(x))'$$ $$g''(x)=c(e^{cx})'+f''(x)$$ But we already prove above that $(e^{cx})'=ce^{cx}$ Therefore, $$g''(x)=c^2e^{cx}+f''(x)$$ *Find $g'(0)$ $$g'(0)=ce^{c\times0}+f'(0)$$ $$g'(0)=c+f'(0)$$ We already know that $f'(0)=5$. So, $$g'(0)=c+5$$ *Find $g''(0)$ $$g''(0)=c^2e^{c\times0}+f''(0)$$ $$g''(0)=c^2+f''(0)$$ We also know that $f''(0)=-2$. $$g''(0)=c^2-2$$ (b) $$h(x)=e^{kx}f(x)$$ *The derivative of $h(x)$ is $$h'(x)=(e^{kx})'f(x)+e^{kx}f'(x)$$ From part a), we can derive $(e^{kx})'=ke^{kx}$. Therefore, $$h'(x)=ke^{kx}f(x)+e^{kx}f'(x)$$ $$h'(x)=e^{kx}(kf(x)+f'(x))$$ *For $x=0$, $$h'(0)=e^{k\times0}(kf(0)+f'(0))$$ From the given information, $f(0)=3$ and $f'(0)=5$. Therefore, $$h'(0)=e^0(3k+5)=3k+5$$ Also for $x=0$, $$h(0)=e^{0\times k}f(0)=1\times3=3$$ *The slope of the tangent line (l) to the curve $h$ where $x=0$ is the value of $h'(0)$. Therefore, the equation of the tangent line (l) is $$(l): y-h(0)=h'(0)(x-0)$$ $$(l): y-3=(3k+5)x$$ $$(l): y=(3k+5)x+3$$
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