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 204: 51


$$y=(\ln2 )x+1$$

Work Step by Step

Re-arranging the function into a differentiable form: $y=e^{(\ln2) x}$ The derivative of the curve is (using the chain rule): $y'=\frac{de^{(\ln2)x}}{d(\ln2)x} \times\frac{d(\ln2)x}{dx}$ $=e^{(\ln 2)x} \times \ln2$ $=\ln 2\times2^x$ Thus, the slope ($m$) of the tangent line at (0,1) would be: $m= y'(0)$ $=\ln 2 \times 2^{(0)}$ (plugging $x=0$ into equation $y'$) $=\ln2$ Thus, plugging in the given points (0,1) into the general point-slope form of a linear equation, an equation of the tangent line is: $y-1=\ln2(x-0)$ Further simplifying the equation gives: $y=(\ln2 )x+1$
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