Real analysis/Differentiation

Definition

We now define the derivative of a function $f$ at a point $a$ to be: $f^{\prime }(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}$.

or equivelantly (exercise!) by

$f^{\prime }(x)=\underset{y\to x}{\lim }\frac{f(y)-f(x)}{y-x}$.

The derivative exists if and only if the limit exists, in which case the function is said to be differentiable at the point $a$. A function is differentiable on an set $A$ if the derivative exists for each $a\in A$. A function could be said to be differentiable if it is differentiable on its whole 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.

Properties

Some properties of the derivative follow immediately from the definition:

Basic Properties

If f and g are differentiable, then:

• $(f+g{)}^{\prime }(x)=f^{\prime }(x)+g^{\prime }(x)$

• $(\lambda f{)}^{\prime }(x)=\lambda f^{\prime }(x)$

Proof

• $(f+g{)}^{\prime }(x)=\underset{y\to x}{\lim }\frac{(f(y)+g(y))-(f(x)+g(x))}{y-x}$

$=\underset{y\to x}{\lim }(\frac{f(y)-f(x)}{y-x}+\frac{g(y)-g(x)}{y-x})$

$=\underset{y\to x}{\lim }\frac{f(y)-f(x)}{y-x}+\underset{y\to x}{\lim }\frac{g(y)-g(x)}{y-x}=f^{\prime }(x)+g^{\prime }(x)$

• $(\lambda f{)}^{\prime }(x)=\underset{y\to x}{\lim }\frac{\lambda f(y)-\lambda f(x)}{y-x}=\lambda \underset{y\to x}{\lim }\frac{f(y)-f(x)}{y-x}=\lambda f^{\prime }(x)$

Theorem(Differentiability Implies Continuity)

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

Proof

Since f is differentiable at x, $\underset{y\to x}{\lim }\frac{f(y)-f(x)}{y-x}=f^{\prime }(x)$.

So $\underset{y\to x}{\lim }[f(y)-f(x)]=\underset{y\to x}{\lim }\frac{f(y)-f(x)}{y-x}\underset{y\to x}{\lim }(y-x)=f^{\prime }(x)0=0$

Thus $\underset{y\to x}{\lim }f(y)=f(x)$, so f is continuous at x.

Theorem(Product Rule)

If f and g are differentiable, then $(fg{)}^{\prime }(x)=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$

Proof

$(fg{)}^{\prime }(x)=\underset{y\to x}{\lim }\frac{f(y)g(y)-f(x)g(x)}{y-x}$ $=\underset{y\to x}{\lim }\frac{f(y)g(y)-f(x)g(y)+f(x)g(y)-f(x)g(x)}{y-x}$

$=\underset{y\to x}{\lim }(\frac{f(y)-f(x))g(y)+f(x)(g(y)-g(x))}{y-x}$

$=\underset{y\to x}{\lim }\frac{f(y)-f(x)}{y-x}\underset{y\to x}{\lim }g(y)+f(x)\underset{y\to x}{\lim }\frac{g(y)-g(x)}{y-x}$$=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$, since g is continuous at x.$◻$

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

$(f\circ g{)}^{\prime }(x)=\underset{y\to x}{\lim }\frac{f(g(y))-f(g(x))}{y-x}$ $=\underset{y\to x}{\lim }\frac{f(g(y))-f(g(x))}{g(y)-g(x)}\frac{g(y)-g(x)}{y-x}$$=\underset{y\to x}{\lim }\frac{f(g(y))-f(g(x))}{g(y)-g(x)}\underset{y\to x}{\lim }\frac{g(y)-g(x)}{y-x}$$=f^{\prime }(g(x))g^{\prime }(x)$

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

Theorem(Chain Rule)

If g is differentiable at x and f is differentiable at g(x), then$(f\circ g{)}^{\prime }(x)=f^{\prime }(g(x))g^{\prime }(x)$

Proof

Define $D_f(y)=\{\begin{array}{cc}\frac{f(y)-f(g(x))}{y-g(x)}& \text{if }y=x\\ f^{\prime }(g(x))& \text{if }y=x\end{array}$

Since f is differentiable, $\underset{y\to x}{\lim }{D}_f(y)=f^{\prime }(x)$, so $D_f(y)$ is continuous at x.

If $g(y)=g(x)$, then $\frac{f(g(y))-f(g(x))}{y-x}$$=\frac{f(g(y))-f(g(x))}{g(y)-g(x)}\frac{g(y)-g(x)}{y-x}$$=D_f(g(y))\frac{g(y)-g(x)}{y-x}$

If $g(y)=g(x)$, then $D_f(g(y))\frac{g(y)-g(x)}{y-x}=f^{\prime }(g(x))0=0=\frac{f(g(y))-f(g(x))}{y-x}$.

Thus $\underset{y\to x}{\lim }\frac{f(g(y))-f(g(x))}{y-x}$$=\underset{y\to x}{\lim }{D}_f(g(y))\frac{g(y)-g(x)}{y-x}$$=\underset{y\to x}{\lim }{D}_f(g(y))\underset{y\to x}{\lim }\frac{g(y)-g(x)}{y-x}$$=f^{\prime }(g(x))g^{\prime }(x)$

Theorem(Rolle's Theorem)

The Extreme Value Theorem easily follows from the fact that the continuous image of a compact set is also compact, and hence bounded and closed within the real numbers, so that it acquires both its maximum and minimum. This Extreme Value Theorem is used to prove the following:

If f(x) is differentiable on [a,b] and f(a)=f(b), then there exists a c within [a,b] such that f'(c)=0.

Proof

Suppose that the function f(x) is constant. Then for all values of c with [a,b], f'(c)=0.

Then suppose that the function f(x) has a value x such that f(x)>f(a). Its maximum f(c) is greater than f(a), and so because it is differentiable, can happen only when f'(c)=0. Thus, c is within [a,b] and f'(c)=0.

Then suppose that the function f(x) has values that are less than or equal to f(a). Then -f(x) has values that are greater than or equal to -f(a), and so has a maximum -f(c), with -f'(c)=0, so f'(c)=0.

Theorem(Mean Value Theorem)

Rolle's theorem can be generalized as follows:

If f(x) and g(x) are differentiable on [a,b] (without both having infinite derivatives at the same point) then there exists a c within [a,b] such that

$\frac{f^{\prime }(c)}{g^{\prime }(c)}=\frac{(f(b)-f(a)}{g(b)-g(a)}$.

Proof

The function h(x) defined by h(x)=f(x)(g(b)-g(a))-g(x)(f(b)-f(a)) has it such that h(a)=h(b), and by Rolle's theorem, there is a c within [a,b] such that h'(c)=0.

Examples

Consider $f:ℝ\to ℝ$ by $f(x)=x$. What is the derivative of $f$ at $a$?

$f^{\prime }(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}=\underset{h\to 0}{\lim }\frac{a+h-a}{h}=\underset{h\to 0}{\lim }\frac{h}{h}=\underset{h\to 0}{\lim }1=1$

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

Similar derivative formula's may also be found.