Example 2 (film)

Consider the function $y(x) = \sqrt{x}$. Determine the equation of the tangent line to the graph of the function at $(4,2)$. In the figure below the situation is sketched.

The general from of a tangent line is $t(x)=ax + b$, with $a$ the slope and $b$ the point of intersection with the $y$-axis. The slope is equal to the derivative of $y(x)$ in $x=4$, hence $a=y'(4)$. At first glance, if might seem impossible to determine the derivative of $y(x)$ by the use of the derivatives of elementary function, given the fact that the square root function is not part of the table, but recall that $y(x) = \sqrt{x} = x^{\tfrac{1}{2}}$. Hence, we can use line (2) from the table.

$$\begin{align} y'(x) &= \tfrac{1}{2} x^{\tfrac{1}{2}-1} = \tfrac{1}{2} x^{-\tfrac{1}{2}} = \dfrac{1}{2}\dfrac{1}{x^{\tfrac{1}{2}}} = \dfrac{1}{2\sqrt{x}},\\ y'(4) &= \dfrac{1}{2\cdot\sqrt{4}} = \tfrac{1}{4}. \end{align}$$
The slope of the tangent line is $\tfrac{1}{4}$, which gives the following equation of the tangent line: $t(x)=\tfrac{1}{4}x+b$. Moreover, we know that the tangent line goes through the point $(x,y)=(4,\sqrt{4})=(4,2)$ as this is the point where the line is tangent to the graph of $y(x)$. By plugging this point into the tangent line, we can determine $b$:
$$\begin{align} t(x) &= \tfrac{1}{4}x + b\\ 2 &= \tfrac{1}{4}\cdot 4 + b = 1+b\\ b&= 1. \end{align}$$
Hence, the equation of the tangent line is given by $t(x) = \tfrac{1}{4} x + 1$.