Chapter 3, Problem 1CC

To determine

To define: The expression for slope of tangent line to the curve $y=f\left(x\right)$ at the point $\left(a,f\left(a\right)\right)$.

Answer to Problem 1CC

The expression for slope of tangent line to the curve $y=f\left(x\right)$ at the point $\left(a,f\left(a\right)\right)$ is $\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$.

Explanation of Solution

Given information:

The curve $y=f\left(x\right)$ and the point $\left(a,f\left(a\right)\right)$.

The expression for slope mof tangent line to the curve $y=f\left(x\right)$ at the point $\left(a,f\left(a\right)\right)$ is $m=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ , provided that the limit exists.

Let $x-a=h$ then $x=a+h$. When limit of x tends to a, the limit of h tends to 0 so another expression for slope m of tangent line to the curve $y=f\left(x\right)$ at the point $\left(a,f\left(a\right)\right)$ is $m=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ , provided that the limit exists.

For example: Slope of the curve $y=9x$ at the point $\left(1,9\right)$ is,

  $\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{9\left(a+h\right)-9a}{h}\\ =\underset{h\to 0}{\lim }\frac{9a+9h-9a}{h}\\ =\underset{h\to 0}{\lim }\frac{9h}{h}\end{array}$

Simplify it further as,

  $m=9$

Therefore, slope of the curve $y=9x$ at the point $\left(1,9\right)$ is 9.