# 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,
$$y = 3 +2(x-1) \text{ or } y = 2x+1$$