## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 4 - Calculating the Derivative - 4.3 The Chain Rule - 4.3 Exercises - Page 225: 46

#### Answer

$y=x+3$

#### 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 $y-f(a)=f^{\prime}(a)(x-a)$ $f(x)=(x^{3}+7)^{2/3}$ To find $f^{\prime}(x)$, we need the chain rule: $f^{\prime}(x)=\displaystyle \frac{2}{3}(x^{3}+7)^{-1/3}\cdot(x^{3}+7)^{\prime}$ $f^{\prime}(x)=\displaystyle \frac{2}{3}(x^{3}+7)^{-1/3}\cdot 3x^{2}$ $\displaystyle \color{blue}{f^{\prime}(x)=\frac{2x^{2}}{(x^{3}+7)^{1/3}}}$ $f(1)=(1+7)^{2/3}=8^{2/3}=4$ $f^{\prime}(1)=\displaystyle \frac{2}{(1+7)^{1/3}}=\frac{2}{2}=1$ Point-slope form for the line passing through $(1,4),$ with slope $1:$ $y-4=1(x-1)$ ... solve for y (to transform to slope-intercept form).... $y=x-1+4$ $y=x+3$

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