The derivative of a function at a given point is the slope of the tangent line to the function at the given point. In other words, the derivative of a function \(f\) at a point \(c\) is the slope of the tangent line to the curve \(f\) at the point \(c\). The derivative of a function is a key concept in science and engineering and has applications, for example, in machine learning.

#### The definition of the derivative of a function

Let \(f\) be a real-valued function definable on a closed real interval \([a,b]\) with \(a < b\). Let \(c\) be a real number between \(a\) and \(b\). By definition, the derivative of the function \(f\) at the point \(c\), denoted by \(f'(c)\), is $$f'(c) = \lim_{x \to c} \frac{f(x) – f(c)}{x-c},$$ provided this limit exists.

Equivalently, the derivative of the function \(f\) at the point \(c\), denoted by \(f'(c)\), is $$f'(c) = \lim_{h \to 0} \frac{f(c+h) – f(c)}{h},$$ provided this limit exists. In such a case, we say that \(f\) is differentiable at \(c\).

The derivative of a real-valued function \(f\) is the real-valued function, denoted by \(f’\), whose value at \(x\) is $$f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h},$$ provided this limit exists. It is clear that the domain of \(f’\) is the set of those points in the domain of \(f\) for which the function \(f\) is differentiable at them. The derivative of a function \(y= f(x)\) is also denoted by \(\displaystyle \frac{df}{dx}\) or \(\displaystyle \frac{dy}{dx}\)(Leibniz’s notation).

**Exercise**. Find the derivative of the function \(f(x) = \sqrt{x}\) at any point in the domain of \(f\).

Let \(x\) be a positive real number. Observe that $$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = $$ $$ \lim_{h \to 0} \frac{\sqrt{x+h} – \sqrt{x}}{h}.$$ If we multiply the nominator and denominator of the above fraction by \(\sqrt{x+h} + \sqrt{x}\), we see that $$f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} = $$ $$\frac{1}{2\sqrt{x}}.$$ Note that the domain of \(f(x) = \sqrt{x}\) is the set of non-negative real numbers and the domain of its derivative is the set of all positive real numbers. It is clear that \(f(x) = \sqrt{x}\) is not differentiable at \(x = 0\). However, the right-hand derivative of \(f(x) = \sqrt{x}\) at \(x=0\) is \(+\infty\).

#### Differentiable functions are continuous

Let us recall that a function \(f\) is continuous at \(c\) if the limit of the function \(f\) at \(c\) exists and $$\lim_{x \to c} f(x) = f(c).$$

**Result**. Any function differentiable at a point is continuous at the point.

The proof is as follows: Let \(f\) be differentiable at a point \(c\). Observe that $$\lim_{x \to c} (f(x) – f(c)) = $$ $$\lim_{x \to c} \frac{f(x) – f(c)}{x-c} \cdot \lim_{x \to c} (x-c) = $$ $$f'(c) \cdot 0 = 0.$$ This implies that $$\lim_{x \to c} f(x) = f(c),$$ and so, \(f\) is continuous at \(c\).

**Remark**. It is very important to note that the converse of the above result is not correct. For example, the absolute value function \(f(x) = \vert x \vert\) is continuous at zero while it is not differentiable at zero (prove this).