Question

State what it means for a scalar function $y=f\left(x\right)$ to be differentiable at a point c.

Short answer

The limit obtained is called the derivate of f at c the process of finding the derivative of a function is a called differentiation

Step-by-step solution

Step 1. Given information

Given scalar function$y=f\left(x\right)$

Step 2. The object is to define the differentiability of a scalar function y=f(x) at a point c.

Let $x=c$be a point in the domain of the function $f\left(x\right)$

If$\underset{h\to 0}{\lim }\frac{f(c+h)-f\left(c\right)}{h}$ exists the we say that$f\left(x\right)$ is derivable at $x=c$and write

$\underset{h\to 0}{\lim }\frac{f(c+h)-f\left(c\right)}{h}=f\text{'}\left(c\right)$or let $x=c$be a point in the domain of the function $f\left(x\right)$

If $\underset{h\to 0}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$exists then we say that $f\left(x\right)$is derivate at $x=c$and $\underset{h\to 0}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=f\text{'}\left(c\right)$

The limit obtained is called the derivate of f at cthe process of finding the derivative of a function is a called differentiation

Step 3. Taking an example

Let $f\left(x\right)=x^2$and$C=2$ in the domain of$f\left(x\right)$

Consider $\underset{h\to 0}{\lim }\frac{f(c+h)-f\left(c\right)}{h}$

$$\begin{gathered} =\underset{h\to 0}{\lim }\frac{f(2+h)-f\left(2\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{4h+h^2}{h} \\ =\underset{h\to 0}{\lim }4+h \\ =4+0 \\ =4 \end{gathered}$$

exists

Thus the function is differentiable at 2