Calculus: Early Transcendentals, Chapter 3, 3.1, Section 3.1, Problem 38

Determine an equation of the tangent line $y = x - \sqrt{x}$ to the curve at the point $(1,0)$.
Graph the curve and the tangent line on the same screen.

Recall that the first derivative is equal to the slope of the tangent line at a certain point. So,

$
\begin{equation}
\begin{aligned}
m = \frac{dy}{dx} &= \frac{d}{dx} \left( x - x^{\frac{1}{2}} \right) \\
\\
&= \frac{d}{dx} (x) - \frac{d}{dx} \left( x^{\frac{1}{2}} \right) \\
\\
&= (1) - \frac{1}{2} x^{\frac{1}{2} - 1}\\
\\
&= 1 - \frac{1}{2} x^{-\frac{1}{2}} \\
\\
&= 1 - \frac{1}{2x^{\frac{1}{2}}}
\end{aligned}
\end{equation}
$


Thus, the slope of the tangent line at point $(1,0)$ is
$\displaystyle m = 1 - \frac{1}{2(1)^{\frac{1}{2}}} = 1 - \frac{1}{2} = \frac{1}{2}$

Therefore, by using the point slope form, the equation of the tangent line will be

$
\begin{equation}
\begin{aligned}
y - y_1 &= m(x - x_1) \\
\\
y - 0 &= \frac{1}{2} (x - 1)\\
\\
y &= \frac{1}{2} x - \frac{1}{2}
\end{aligned}
\end{equation}
$

And the graph is