Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.3 The Chain Rule - 4.3 Exercises - Page 225: 47



Work Step by Step

The slope of the tangent at x=a is the derivative of f(x), calculated in x=a. The tangent line has slope $f^{\prime}(a)$, and passes through $(a,f(a)).\\\\$ The point-slope equation is $y-f(a)=f^{\prime}(a)(x-a)$ We find $f'(x):$ $f(x)=x(x^{2}-4x+5)^{4}=u(x)\cdot v(x)\qquad $(... use the product rule) $u(x)=x\qquad u^{\prime}(x)=1$ $v(x)=(x^{2}-4x+5)^{4}\qquad $(... we use the chain rule) $v^{\prime}(x)=4(x^{2}-4x+5)^{3}\cdot(x^{2}-4x+5)^{\prime}\\=4(x^{2}-4x+5)^{3}\cdot(2x-4)$ $f^{\prime}(x)=u(x)\cdot v^{\prime}(x)+v(x)\cdot u^{\prime}(x)$. $=x\cdot 4(x^{2}-4x+5)^{3}\cdot(2x-4) +1\cdot(x^{2}-4x+5)^{4}$ $=(x^{2}-4x+5)^{3}\cdot [4x(2x-4)+(x^{2}-4x+5)]$ $\color{blue}{f^{\prime}(x)=(x^{2}-4x+5)^{3}(9x^{2}-20x+5)}$ $f(2)=x(4-4(2)+5)^{4}=2(1)^{4}=2$ $f^{\prime}(2)=(4-4(2)+5)^{3}(9(4)-20(2)+5)=(1)^{3}(1)=1$ Point-slope form for the line passing through $(2,2),$ with slope $1:$ $y-2=1(x-2)$ ... solve for y (to transform to slope-intercept form)... $y-2=x-2$ $y=x-2+2$ $y=x$
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