Real Analysis/Differentiation

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Real Analysis
Differentiation

Definition[edit]

We are now ready to define the derivative of a function

Let f:\mathbb{R}\to\mathbb{R}

Let a\in\mathbb{R}

We say that ƒ(x) is differentiable at x=a if and only if there exists a real number L such that

\lim_{x \rightarrow a} {f(x)-f(a) \over x-a}=L.

L is said to be the derivative of ƒ at a and is denoted by f'(a)

A function is said to be differentiable on a set A if the derivative exists for each a in A. A function is differentiable if it is differentiable on its entire domain.

Conceptually, finding the derivative means finding the slope of the tangent line to the function. Thus the derivative can be thought of as a linear, or first-order, approximation of the function.

Properties[edit]

Some properties of the derivative follow immediately from the definition:

Basic Properties[edit]

If f and g are differentiable, then:


  •  (f+g)'(x) =  f'(x) + g'(x)
  •  (\lambda f)'(x) = \lambda f'(x)

Proof[edit]

  • (f+g)'(x) = \lim_{y \rightarrow x} {(f(y)+ g(y))-(f(x)+g(x)) \over y-x}

= \lim_{y \rightarrow x} \left ( {f(y) - f(x) \over y-x} + {g(y) - g(x) \over y-x} \right )

 = \lim_{y \rightarrow x} {f(y) - f(x) \over y-x} + \lim_{y \rightarrow x} {g(y) - g(x) \over y-x} = f'(x) + g'(x)


  •  ( \lambda f)'(x) = \lim_{y \rightarrow x} { \lambda f(y) - \lambda f(x) \over y-x} = \lambda \lim_{y \rightarrow x} {f(y) - f(x) \over y-x} = \lambda f'(x)

Theorem(Differentiability Implies Continuity)[edit]

If f is differentiable at x, it is continuous at x

Proof[edit]

Since f is differentiable at x, \lim_{y \rightarrow x} {f(y)-f(x) \over y-x} = f'(x).

So \lim_{y \rightarrow x} \left[ f(y)-f(x) \right] = \lim_{y \rightarrow x} {f(y)-f(x) \over y-x} \lim_{y \rightarrow x} \left( y-x \right) = f'(x) \; 0 = 0

Thus \lim_{y \rightarrow x} f(y) = f(x), so f is continuous at x.

Theorem(Product Rule)[edit]

If f and g are differentiable, then  (fg)'(x) = f'(x)g(x) + f(x)g'(x)

Proof[edit]

 (fg)'(x) = \lim_{y \rightarrow x} {f(y)g(y)-f(x)g(x) \over y-x} = \lim_{y \rightarrow x} {f(y)g(y)-f(x)g(y)+f(x)g(y)-f(x)g(x) \over y-x}

 = \lim_{y \rightarrow x} ({f(y)-f(x))g(y)+f(x)(g(y)-g(x)) \over y-x}

= \lim_{y \rightarrow x} {f(y)-f(x) \over y-x}\lim_{y \rightarrow x} g(y) + f(x) \lim_{y \rightarrow x} {g(y)-g(x) \over y-x}  = f'(x)g(x) + f(x)g'(x), since g is continuous at x.\Box


The following theorem is a bit trickier to prove than it seems. We would like to use the following argument:


(f \circ g)'(x) = \lim_{y \rightarrow x} {f(g(y))-f(g(x)) \over y-x} = \lim_{y \rightarrow x} {f(g(y))-f(g(x)) \over g(y)-g(x)}{g(y)-g(x) \over y-x} = \lim_{y \rightarrow x} {f(g(y))-f(g(x)) \over g(y)-g(x)}\lim_{y \rightarrow x}{g(y)-g(x) \over y-x} = f'(g(x))g'(x)


The problem is that g(y) - g(x) may be zero at points arbitrarily close to x, and therefore {f(g(y))-f(g(x)) \over g(y)-g(x)} would not be continuous at these points. Thus we apply a clever lemma as follows:

Theorem (Caratheodory's Lemma)[edit]

Let f:\mathbb{R}\to\mathbb{R}

We say that f(x) is differentiable at x=c if and only if there exists a continuous function \phi :\mathbb{R}\to\mathbb{R} that satisfies

(x-c)\phi (x)=f(x)-f(c)\forall x\in\mathbb{R}

Proof[edit]

(\Rightarrow)Let f(x) be differentiable at x=c and define function \phi :\mathbb{R}\to\mathbb{R} such that

\phi (x)=\frac{f(x)-f(c)}{x-c} for x\neq c and

\phi (c)=f'(c)

It is easy to see that \phi (x) is continuous and that it satisfies the required condition.

(\Leftarrow) Let \phi (x) be a continuous function satisfying (x-c)\phi (x)=f(x)-f(c)\forall x\in\mathbb{R}

For all x\neq c, we have that \phi (x)=\tfrac{f(x)-f(c)}{x-c}

As \phi is continuous, \phi (c)=\lim_{x\to c}\phi (x), that is,

\phi (c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c} which implies that f(x) is differentiable at x=c

Theorem(Chain Rule)[edit]

Let g:\mathbb{R}\to\mathbb{R} be differentiable at c\in\mathbb{R}

Let f:\mathbb{R}\to\mathbb{R} be differentiable at d=g(c)

Let the domain of f be a subset of the image of g.

Then
(i)(f\circ g)(x) is differentiable at x=c

(ii)(f\circ g)'(c)=f'(g(c))g'(c)

Proof[edit]

Caratheodory's Lemma implies that there exist continuous functions \phi ,\gamma :\mathbb{R}\to\mathbb{R} such that (x-c)\gamma (x)=g(x)-g(c) and (g(x)-g(c))\phi (g(x))=f(g(x))-f(g(c))

Now, consider the function \eta (x)=\phi (g(x))\gamma (x). Obviously, \eta (x) is continuous.

Also, it satisfies (x-c)\eta (x)=(f\circ g)(x)-(f\circ g)(c). Hence, by Caratheodory's Lemma, (f\circ g)(x) is differentiable at x=c and that (f\circ g)'(c)=\eta (c)=f'(g(c))g'(c)

Examples[edit]

Consider  f: \mathbb{R} \rightarrow \mathbb{R} by  f(x) = x . What is the derivative of  f at  a ?

  
f'(a) = \lim_{h \rightarrow 0} \frac{f(a+h) -f(a)}{h} = \lim_{h \rightarrow 0} \frac{a+h - a}{h} = \lim_{h \rightarrow 0} \frac{h}{h} = \lim_{h \rightarrow 0} 1 = 1

So, here we see that  f'(a) =1 . Since  a was an arbtarily choosen point we conculde that  f'(a) = 1  \forall a \in \mathbb{R}

Similar derivative formula's may also be found.

Exercises[edit]

  • Find the derivatives of the common functions: Polynomial, Trigonometric, Exponential and Logarithmic.
  • Some of the most popular counter examples to illustrate properties of continuity and differentiability are functions involving f(x)=\sin ( \tfrac{1}{x} )

    (i)Prove that f(x)=\sin ( \tfrac{1}{x} )\forall x\neq 0, f(0)=0 is not continuous at x=0

    (ii)Prove that the function f(x)=x\sin (\tfrac{1}{x} )\forall x\neq 0, f(0)=0 is continuous but not differentiable at x=0

    (iii)Prove that f(x)=x^2 \sin( \tfrac{1}{x} )\forall x\neq 0, f(0)=0 is differentiable at x=0