## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 3 - Differentiation - 3.1 Definition of the Derivative - Exercises - Page 103: 44

#### Answer

$$2$$ $$y=2 x+3$$

#### Work Step by Step

Since \begin{aligned} f^{\prime}(-1)&=\lim _{h \rightarrow 0} \frac{f(-1+h)-f(-1)}{h}\\ &=\lim _{h \rightarrow 0} \frac{\frac{1}{1+h^{2}-2 h}-1}{h}\\ &=\lim _{h \rightarrow 0} \frac{1-1-h^{2}+2 h}{h\left(1+h^{2}-2 h\right)}\\ &=\lim _{h \rightarrow 0} \frac{-h^{2}+2 h}{h\left(1+h^{2}-2 h\right)}\\ &=\lim _{h \rightarrow 0} \frac{-h+2}{1+h^{2}-2 h}\\ &=2 \end{aligned} Thus at $a=-1,$ the equation of the tangent line is $y-f(-1)=f^{\prime}(-1)(x-(-1))$ \begin{align*} y-1&=2(x+1)\\ y&=1+2 x+2\\ y&=2 x+3 \end{align*}

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