2.2 | The Derivative at a Point


$$
\begin{array}{c}
\text{Average rate of change of } f \\
\text{over the interval from } a \text{ to } a+h
\end{array}
= \frac{f(a+h) -f(a)}{h}.
$$

The derivative of \(f\) at \(a\), written \(f'(a)\), is defined as
$$
\begin{array}{c}
\text{Rate of change} \\
\text{ of } f \text{ at } a
\end{array}
= f'(a) = \lim_{h \rightarrow 0} \frac{f(a+h) -f(a)}{h}.
$$
If the limit exists, then \(f\) is said to be differentiable at \(a\).
Find the slope of the tangent line to the parabola \(y = 4x-x^2\) at the point \((1,3)\) using the limit definition. Use this information to find an equation for the tangent line to the graph of the parabola at \((1,3)\).


Let \(f(x) = 4x-x^2\), so that
$$
\begin{aligned}
f'(1) & = \lim_{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} = \lim_{h \rightarrow 0} \frac{4(1+h)-(1+h)^2 – (4(1)-1^2)}{h} \\
& = \lim_{h \rightarrow 0} \frac{4+4h-1-2h-h^2 – 3}{h} = \lim_{h \rightarrow 0} \frac{2h-h^2}{h} \\
& = \lim_{h \rightarrow 0} \frac{h(2-h)}{h} = \lim_{h \rightarrow 0} (2-h) = 2 \\
\end{aligned}
$$
So, the slope of the tangent line to the parabola \(y = 4x-x^2\) at \((1,3)\) is 2.

An equation for the tangent line to the parabola at \((1,3)\) is given by
$$
y = f(a) +f'(a)(x-a)
$$
where \(f(a) = 3\), \(f'(a) = 2\), and \(a = 1\). That is,
$$
y = 3 +2(x-1) \text{ or } y = 2x+1
$$

E 2.2 Exercises