# Definition of Derivative

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There is one limit that is used very frequently in applications of calculus and different sciences. This limit has form lim_(h->0)(f(x+h)-f(x))/h and has special notation.

Definition. The derivative of function f is a function that is denoted by f'(x) and calculated as f'(x)=lim_(h->0)(f(x+h)-f(x))/h.

Process of finding derivative is called differentiating.

There are different notations for derivative.

Leibniz notation: (dy)/(dx), (df)/(dx), d/(dx)f(x). Note, that here (dy)/(dx) is not ratio - it is just synonym for f'(x). It is very useful and suggestive notation, derivative is defined as a limit of ratio. If we want to indicate the value of a derivative in Leibniz notation at a specific number a, we use the notation (dy)/(dx)|_(x=a).

Lagrange notation: y', f'(x).

Cauchy notation: Dy, Df(x).

In Lagrange and Cauchy notations if we need to explicitly state with respect to what variable we want to take derivative, we write y'_x, f'_x, D_xy, D_x f.

We will mainly use Lagrange notation.

Example 1. If f(x)=x^2, find f'(x) and f(3).

We have that f(x+h)=(x+h)^2.

Thus,

f'(x)=lim_(h->0)(f(x+h)-f(x))/h=lim_(h->0)((x+h)^2-(x)^2)/(h)=lim_(h->0)(x^2+2xh+h^2-x^2)/h=

=lim_(h->0)(2xh+h^2)/h=lim_(h->0)(2x+h)=2x.

So, f'(x)=2x and f'(3)=2*3=6.

Example 2. Find f'(x) if f(x)=sqrt(x) and state domain of f'.

f'(x)=lim_(h->0)(sqrt(x+h)-sqrt(x))/h=lim_(h->0)(sqrt(x+h)-sqrt(x))/h (sqrt(x+h)+sqrt(x))/(sqrt(x+h)+sqrt(x))=lim_(h->0)(x+h-x)/(h(sqrt(x+h)+sqrt(x)))=

=lim_(h->0)1/(sqrt(x+h)+sqrt(x))=1/(sqrt(x+0)+sqrt(x))=1/(2sqrt(x))

Thus, domain of f' is (0,oo). As can be seen it is smaller then domain of f(x)=sqrt(x).

What if function is given by table of values?

Example 3. Estimate f'(1) and f'(1.5) if

 x 1 1.5 2 f(x) 2 3 7

We have that f'(x)=lim_(h->0)(f(x+h)-f(x))/h. But if we take h=0.5 we will obtain following approximation: f'(x)~~(f(x+0.5)-f(x))/0.5=2(f(x+0.5)-f(x)).

Thus,

f'(1)~~2(f(1+0.5)-f(1))=2(f(1.5)-f(1))=2(3-2)=2.

f'(1.5)~~2(f(1.5+0.5)-f(1.5))=2(f(2)-f(1.5))=2(7-3)=8.

Definition. Function f is differentiable at a if f'(a) exists.

Definition. Function f is differentiable on open interval (a,b) or (a,oo) or (-oo,a) or (-oo,oo) if it is differentiable at every number in that interval.

Example 4. Where is f(x)=|x| is differentiable?

If x>0, then |x|=x and we can choose h small enough so that x+h>0 and hence |x+h|=x+h.

Therefore, for x>0 f'(x)=lim_(h->0)(|x+h|-|x|)/h=lim_(h->0)(x+h-x)/h=lim_(h->0)1=1 and so f is differentiable for any x>0.

Similarly, if x<0, then |x|=-x and we can choose h small enough so that x+h<0 and hence |x+h|=-(x+h).

Therefore, for x<0 f'(x)=lim_(h->0)(|x+h|-|x|)/h=lim_(h->0)(-(x+h)-(-x))/h=lim_(h->0)-1=-1 and so f is differentiable for any x<0.

For x=0 we have to investigatef'(0)=lim_(h->0)(|0+h|-|0|)/h=lim_(h->0)(|h|)/h.

Let's compute one-sided limits:

lim_(h->0^+)|h|/h=lim_(h->0^+)h/h=lim_(h->0^+)1=1

lim_(h->0^-)|h|/h=lim_(h->0^-)-h/h=lim_(h->0^-)-1=-1

Since lim_(h->0^+)(|h|)/h!=lim_(h->0^-)(|h|)/h thenlim_(h->0)(|h|)/h doesn't exist and thus f(x) is not differentiable at x=0.

Therefore, |x| is differentiable at all x except 0.

Theorem. If f is differentiable at a then f is continuous at a. Converse is not true.

Proof. Since f is differentiable at a then f'(a)=lim_(h->0)(f(a+h)-f(a))/h.

If we make substitution x=a+h then h=x-a. As h->0 x->a.

Therefore, f'(a)=lim_(x->a)(f(x)-f(a))/(x-a).

Now consider identity f(x)-f(a)=f(x)-f(a) or equivalently f(x)-f(a)=(f(x)-f(a))/(x-a)(x-a). Taking limits of both sides gives lim_(x->a)(f(x)-f(a))=lim_(x->a)((f(x)-f(a))/(x-a)(x-a)).

Using property of limit of product we obtain thatlim_(x->a)(f(x)-f(a))=lim_(x->a)(f(x)-f(a))/(x-a)lim_(x->a)(x-a) or lim_(x->a)(f(x)-f(a))=f'(a)*(a-a).

Thus, lim_(x->a)(f(x)-f(a))=0.

Now, lim_(x->a)f(x)=lim_(x->a)(f(x)-f(a)+f(a))=

=lim_(x->a)(f(x)-f(a))+lim_(x->a)(f(a))=0+f(a)=f(a).

Therefore lim_(x->a)f(x)=f(a) which means that function is continuous at a.

Converse is not true: if f is continuous at a then it doesn't mean that f is differentiable at a. Consider function f(x)=|x|. It is continuous at x=0, but as was shown in example 4 it is not differentiable at 0.