Find an equation of the tangent line to the curve at the given point. y=x(81,9)

Step 1
Consider the given function:
${\displaystyle y=\sqrt{x}}$
Step 2
Now differentiate the function to find the slope.
${\displaystyle \frac{dy}{dx}=\frac{1}{2\sqrt{x}}}$
Step 3
Now find the slope at given point (81,9).
${\displaystyle m=\frac{dy}{{dx}_{(81,9)}}=\frac{1}{2\sqrt{81}}}$
${\displaystyle =\frac{1}{18}}$
Step 4
Write the equation for a line.
${\displaystyle y-y_1=m(x-x_1)}$
Step 5
Put the values of known quantities.
${\displaystyle y-9=\frac{1}{18}(x-81)}$
${\displaystyle 18y-162=x-81}$
${\displaystyle x-18y+81=0}$

Step 1
Let ${\displaystyle f\left(x\right)=\sqrt{x}}$
Then, by Eq. 3, the slope of the tangent at ${\displaystyle (1,1)}$ is
${\displaystyle m=\underset{h\Rightarrow 0}{lim}\frac{f(81+h)-f\left(9\right)}{h}=\underset{h\Rightarrow 0}{lim}\frac{\sqrt{81+h}-9}{h}}$
${\displaystyle =\underset{h\Rightarrow 0}{lim}\frac{(\sqrt{81+h}-9)(\sqrt{81+h}+9)}{h(\sqrt{81+h}+9)}=\underset{h\Rightarrow 0}{lim}\frac{81+h-9}{h(\sqrt{81+h}+9)}}$
${\displaystyle =\underset{h\Rightarrow 0}{lim}\frac{h}{h(\sqrt{81+h}+9)}=\underset{h\Rightarrow 0}{lim}\frac{1}{\sqrt{81+h}+9}=\frac{1}{18}}$
Using the point-slope form of the equation of a line, we find that an equation of the tangent at the point ${\displaystyle (1,1)}$ is
${\displaystyle y-9=\frac{1}{18}(x-81)}$
Which simplifies to ${\displaystyle x-18y+81=0}$