## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 4 - Calculating the Derivative - 4.3 The Chain Rule - 4.3 Exercises - Page 225: 45

#### Answer

$y=\displaystyle \frac{3}{5}x+\frac{16}{5}$

#### 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 $y-f(a)=f^{\prime}(a)(x-a)$ To find $f^{\prime}(x)$, we need the chain rule: $f(x)=(x^{2}+16)^{1/2}=[w(x)]^{1/2}$ $f^{\prime}(x)=\displaystyle \frac{1}{2}[w(x)]^{-1/2}\cdot w^{\prime}(x)$ $f^{\prime}(x)=\displaystyle \frac{1}{2}(x^{2}+16)^{-1/2}\cdot 2x=\color{blue}{\frac{x}{\sqrt{x^{2}+16}}}$ $f(3)=\sqrt{9+16}=\sqrt{25}=5,$ $f^{\prime\prime}(3)=\displaystyle \frac{3}{\sqrt{9+16}}=\frac{3}{5}$ Point-slope form for the line passing through $(3,5)$, with slope $\displaystyle \frac{3}{5}$ $y-5=\displaystyle \frac{3}{5}(x-3)$ ... solve for y (to transform to slope-intercept form)... $y-5=\displaystyle \frac{3}{5}x-\frac{9}{5}$ $y=\displaystyle \frac{3}{5}x-\frac{9}{5}+5$ $y=\displaystyle \frac{3}{5}x+\frac{16}{5}$

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