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Section 3.1 Tangents and the Derivative at a Point

Definition 3.1.1.

The slope of the curve \(y=f(x)\) at the point \((x_0,f(x_0))\) is

\begin{equation*} \lim_{h\rightarrow 0} \frac{f(x_0+h)-f(x_0)}{h} \end{equation*}

provided the limit exists. This is also called the instantaneous rate of change.

The tangent line to the curve is the line with that slope through the point.

Example 3.1.2.

Find the equation of the tangent line to \(f(x)=-2\sqrt{x+1}\) at \(x=3\text{.}\)

Solution.

First, the slope using the definition:

\begin{align*} \lim_{h\to 0} \frac{f(x_0+h)-f(x_0)}{h} \amp = \lim_{h\to 0} \frac{f(3+h)-f(3)}{h}\\ \amp = \lim_{h\to 0} \frac{(-2\sqrt{(3+h)+1})-(-2\sqrt{3+1})}{h}\\ \amp = \lim_{h\to 0} \frac{-2\sqrt{h+4}+4}{h}\\ \amp = \lim_{h\to 0} \frac{-2\sqrt{h+4}+4}{h}\frac{-2\sqrt{h+4} - 4}{-2\sqrt{h+4}-4}\\ \amp = \lim_{h\to 0} \frac{4(h+4)-16}{h(-2\sqrt{h+4}-4)}\\ \amp = \lim_{h\to 0} \frac{4h}{h(-2\sqrt{h+4}-4)}\\ \amp = \lim_{h\to 0} \frac{4}{-2\sqrt{h+4}-4}\\ \amp = \frac{4}{-2\sqrt{0+4}-4}\\ \amp = \frac{-1}{2} \end{align*}

Compute \(f(3) = -2\sqrt{3+1} = -4\text{,}\) so \((3,-4)\) is a point on the curve. Use the point-slope form of the line.

\begin{align*} y-y_0 \amp = m(x-x_0)\\ y-(-4) \amp = \frac{-1}{2}(x-3)\\ y + 4 \amp = \frac{-1}{2}x + \frac{3}{2}\\ y \amp = \frac{-1}{2}x - \frac{5}{2} \end{align*}

Definition 3.1.3.

The derivative of a function \(f\) at a point \(x_0\) is

\begin{equation*} f'(x_0)=\lim_{h\rightarrow 0} \frac{f(x_0+h)-f(x_0)}{h} \end{equation*}

provided the limit exists.

The following are all interpretations of \(\displaystyle\lim_{h\rightarrow 0} \frac{f(x_0+h)-f(x_0)}{h}\text{:}\)

  • The slope of the graph of \(y=f(x)\) at \(x_0\text{.}\)

  • The slope of the tangent line to the curve \(y=f(x)\) at \(x_0\text{.}\)

  • The instantaneous rate of change of \(f(x)\) with respect to \(x\) at \(x_0\text{.}\)

  • The derivative at \(x_0\text{,}\) i.e., \(f'(x_0)\text{.}\)