Question

differentiate the function. then find an equation of the tangent line at the indicated point on the graph of the function.$y = f(x) = \frac{5}{\sqrt{x-2}}, \quad (x,y)=(3,5)$the derivative of the function $y = f(x) = \frac{5}{\sqrt{x-2}}$ is $-\frac{5}{2\sqrt{x-2}(x-2)}$(type an exact answer, using radicals as needed.)an equation of the tangent line is $\square$.(type an equation.)

Explanation:

Step1: Rewrite function for differentiation

Rewrite $f(x)=\frac{5}{\sqrt{x-2}}$ as $f(x)=5(x-2)^{-\frac{1}{2}}$

Step2: Find derivative (given)

$f'(x)=-\frac{5}{2\sqrt{x-2}(x-2)}$ or simplified: $f'(x)=-\frac{5}{2(x-2)^{\frac{3}{2}}}$

Step3: Compute slope at $x=3$

Substitute $x=3$ into $f'(x)$:
$f'(3)=-\frac{5}{2(3-2)^{\frac{3}{2}}}=-\frac{5}{2(1)^{\frac{3}{2}}}=-\frac{5}{2}$

Step4: Use point-slope form

Point-slope formula: $y-y_1=m(x-x_1)$, where $(x_1,y_1)=(3,5)$ and $m=-\frac{5}{2}$
$y-5=-\frac{5}{2}(x-3)$

Step5: Simplify to slope-intercept form

Expand and rearrange:
$y-5=-\frac{5}{2}x+\frac{15}{2}$
$y=-\frac{5}{2}x+\frac{15}{2}+5$
$y=-\frac{5}{2}x+\frac{15}{2}+\frac{10}{2}=-\frac{5}{2}x+\frac{25}{2}$

Answer:

The derivative is $-\frac{5}{2(x-2)^{\frac{3}{2}}}$
An equation of the tangent line is $y=-\frac{5}{2}x+\frac{25}{2}$ (or in standard form: $5x+2y=25$)