Question

consider the following. (the graph is called a witch of agnesi.)
find the derivative of the function.
$f(x)=\frac{\left((9)(2x)-(x^2 + 5)\
ight)}{(x^2 + 5)^2}$
find $f(x)$ when $x = 2$
$f(2)= 36/81$
give the slope of the graph of the function at the point $(2, 1)$.
slope =
find an equation of the tangent line to the graph of $f$ at the point $(2, 1)$.
$y = $

Explanation:

Step1: Apply quotient rule

The quotient rule for $\frac{u}{v}$ is $\frac{u'v - uv'}{v^2}$. Here $u=9$, $u'=0$; $v=x^2+5$, $v'=2x$.
$$f'(x)=\frac{0 \cdot (x^2+5) - 9 \cdot 2x}{(x^2+5)^2}=\frac{-18x}{(x^2+5)^2}$$

Step2: Substitute $x=2$ into $f'(x)$

Calculate denominator first: $(2^2+5)^2=(4+5)^2=81$. Numerator: $-18 \cdot 2=-36$.
$$f'(2)=\frac{-36}{81}=-\frac{4}{9}$$

Step3: Identify slope at $(2,1)$

The slope of the graph at a point is $f'(x)$ at that $x$.
$$\text{slope}=f'(2)=-\frac{4}{9}$$

Step4: Find tangent line equation

Use point-slope form $y - y_1 = m(x - x_1)$, where $(x_1,y_1)=(2,1)$, $m=-\frac{4}{9}$.
$$y - 1 = -\frac{4}{9}(x - 2)$$
Simplify: $y = -\frac{4}{9}x + \frac{8}{9} + 1 = -\frac{4}{9}x + \frac{17}{9}$

Answer:

$f'(x)=\frac{-18x}{(x^2+5)^2}$
$f'(2)=-\frac{4}{9}$
$\text{slope}=-\frac{4}{9}$
$y=-\frac{4}{9}x+\frac{17}{9}$