Calculus with Applications (10th Edition)

Published by Pearson

Chapter 4 - Calculating the Derivative - 4.4 Derivatives of Exponential Functions - 4.4 Exercises - Page 232: 31

Answer

$\displaystyle \frac{(1-t)e^{3t}-4e^{2t}+(1+t)e^{t}}{(e^{2t}+1)^{2}}$

Work Step by Step

$y=\displaystyle \frac{te^{t}+2}{e^{2t}+1}=\frac{u(t)}{v(t)}$ $\begin{array}{ll} & \text{...Apply quotient rule, } \left[\dfrac{u(x)}{v(x)}\right]^{\prime} =\dfrac{v(x)\cdot u^{\prime}(x)-u(x)\cdot v^{\prime}(x)}{[v(x)]^{2}}.\\\\ \dfrac{dy}{dt} & =\dfrac{(e^{2t}+1)(te^{t}+2)^{\prime}-(te^{t}+2)(e^{2t}+1)^{\prime}}{(e^{2t}+1)^{2}}\\ & \text{...apply product rule, }\ \ [te^{t}]'=te^{t}+e^{t}\cdot 1\\\\ & \text{... and } \quad[e^{2t}]'=e^{2t}\cdot2 \\ & \\\\ & =\dfrac{(e^{2t}+1)(te^{t}+e^{t}\cdot 1)-(te^{t}+2)(2e^{2t})}{(e^{2t}+1)^{2}} \\\\ & =\dfrac{(e^{2t}+1)(te^{t}+e^{t})-(te^{t}+2)(2e^{2t})}{(e^{2t}+1)^{2}} \\\\ & =\dfrac{te^{3t}+e^{3t}+te^{t}+e^{t}-2te^{3t}-4e^{2t}}{(e^{2t}+1)^{2}} \\\\ & =\dfrac{-te^{3t}+e^{3t}+te^{t}+e^{t}-4e^{2t}}{(e^{2t}+1)^{2}} & \\\\ & =\dfrac{(1-t)e^{3t}-4e^{2t}+(1+t)e^{t}}{(e^{2t}+1)^{2}} \end{array}$

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