There is one limit that is used very frequently in the applications of calculus and different sciences. This limit has the form limh→0hf(x+h)−f(x) and has a special notation.
Definition. The derivative of a function f is a function that is denoted by f′(x) and is calculated as f′(x)=limh→0hf(x+h)−f(x).
The process of finding the derivative is called differentiating.
There are different notations for the derivative.
Leibniz's notation:dxdy, dxdf, dxdf(x). Note that here dxdy is not a ratio, it is just a synonym for f′(x). It is a very useful and suggestive notation, the derivative is defined as the limit of the ratio. If we want to indicate the value of the derivative in Leibniz's notation at a specific point a, we use the notation dxdy∣x=a.
Lagrange's notation:y′, f′(x).
Cauchy's notation:Dy, Df(x).
In Lagrange's and Cauchy's notations, if we need to state explicitly with respect to what variable we take the derivative, we write yx′, f′x, Dxy, Dxf.
Definition. A function f is differentiable at a if f′(a) exists.
Definition. A function f is differentiable on an open interval (a,b) or (a,∞), or (−∞,a), or (−∞,∞) if it is differentiable at every number in that interval.
Example 4. Where is f(x)=∣x∣ differentiable?
If x>0, we can say that ∣x∣=x and we can choose an h small enough, so that x+h>0, and, hence, ∣x+h∣=x+h.
Therefore, for x>0, f′(x)=limh→0h∣x+h∣−∣x∣=limh→0hx+h−x=limh→01=1, and so, f is differentiable for any x>0.
Similarly, if x<0, we can say that ∣x∣=−x and we can choose an h small enough, so that x+h<0, and, hence, ∣x+h∣=−(x+h).
Therefore, for x<0, f′(x)=limh→0h∣x+h∣−∣x∣=limh→0h−(x+h)−(−x)=limh→0−1=−1, and so, f is differentiable for any x<0.
For x=0, we have to investigatef′(0)=limh→0h∣0+h∣−∣0∣=limh→0h∣h∣.
Since limh→0+h∣h∣=limh→0−h∣h∣, we can state that limh→0h∣h∣ doesn't exist and thus f(x) is not differentiable at x=0.
Therefore, ∣x∣ is differentiable at all values of x except 0.
Theorem. If f is differentiable at a, f is continuous at a. The converse is not true.
Proof. Since f is differentiable at a, we can say that f′(a)=limh→0hf(a+h)−f(a).
If we make the substitution x=a+h, we have that h=x−a. As h→0, x→a.
Therefore, f′(a)=limx→ax−af(x)−f(a).
Now, consider the identity f(x)−f(a)=f(x)−f(a), or, equivalently, f(x)−f(a)=x−af(x)−f(a)(x−a). Taking the limits of both sides gives limx→a(f(x)−f(a))=limx→a(x−af(x)−f(a)(x−a)).
Using the property of the limit of a product, we obtain thatlimx→a(f(x)−f(a))=limx→ax−af(x)−f(a)limx→a(x−a), or limx→a(f(x)−f(a))=f′(a)⋅(a−a).
Thus, limx→a(f(x)−f(a))=0.
Now, limx→af(x)=limx→a(f(x)−f(a)+f(a))=
=limx→a(f(x)−f(a))+limx→a(f(a))=0+f(a)=f(a).
Therefore, limx→af(x)=f(a), which means that the function is continuous at a.
The converse is not true: if f is continuous at a, it doesn't mean that f is differentiable at a. Consider the function f(x)=∣x∣. It is continuous at x=0, but, as was shown in example 4, it is not differentiable at 0.