# Chapter 3 - Differentiation - 3.1 Definition of the Derivative - Exercises - Page 103: 28

$y=-\frac{1}{54}x+\frac{1}{2}$

#### Work Step by Step

Given that $f(x)=\frac{1}{\sqrt x}$ and $a=9$. $f(a)=f(9)=\frac{1}{\sqrt {9}}=\frac{1}{3}$ $f'(a)=f'(9)=\lim\limits_{x \to 9}\frac{f(x)-f(9)}{x-9}=\lim\limits_{x \to 9}\frac{\frac{\sqrt 9-\sqrt x}{\sqrt 9\sqrt x}\times\frac{\sqrt 9+\sqrt x}{\sqrt 9+\sqrt x}}{x-9}$ $=\lim\limits_{x \to 9}\frac{\frac{9-x}{9\sqrt x+x\sqrt 9}}{x-9}=\lim\limits_{x \to 9}\frac{\frac{-(x-9)}{9\sqrt {x}+x\sqrt {9}}}{x-9}=\lim\limits_{x \to 9}\frac{-1}{9\sqrt {x}+x\sqrt 9}$ $=\frac{-1}{9\sqrt 9+9\sqrt 9}=-\frac{1}{54}$ The equation of the tangent line in the point-slope form is $y-f(a)=f'(a)(x-a)$ In this case, $y-\frac{1}{3}=-\frac{1}{54}(x-9)$ Or $y=\frac{1}{3}+\frac{9}{54}-\frac{x}{54}=-\frac{1}{54}x+\frac{1}{2}$

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