The Derivative
Next we talk about differentialbility
Let I be any interval on \mathbb{R}, and let c \in I. We say that f \colon I \to \mathbb{R} is differentiable at c if
\lim_{x \to c} \frac{f(x)-f(c)}{x-c}
exists and is finite.
If f is differentiable at every point in a subset S \subseteq I, we say that f is differentiable on S. If f has a derivative at c, we notate f^\prime(c).
Let f \colon \mathbb{R} \to \mathbb{R} be a constant function. That is f(x) = d for some d \in \mathbb{R}.
Claim: The derivative of f is zero at every point c \in \mathbb{R}.
We use out formula for differentiability
f^\prime(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} = \lim_{x \to c} \frac{d - d}{x - c} = 0
Let f \colon \mathbb{R} \to \mathbb{R} be the identity function. That is f(x) = x for all x \in \mathbb{R}.
Claim: The derivative of f is one at every point c \in \mathbb{R}.
We use out formula for differentiability
f^\prime(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} = \lim_{x \to c} \frac{x - c}{x - c} = 1
Let f \colon \mathbb{R} \to \mathbb{R} where f(x) = x^2.
Claim: The derivative of f is 2c at every point c \in \mathbb{R}.
We use our formula for differentiability
f^\prime(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} = \lim_{x \to c} \frac{x^2 - c^2}{x - c}
Using the formula x^2 - y^2 = (x-y)(x+y), we get
f^\prime(c) = \lim_{x \to c} \frac{(x-c)(x+c)}{x - c} = \lim_{x \to c} x + c = 2c
We introduce another theorem
Let I be an interval containing c and f\colon I \to \mathbb{R}. Then f is differentiable at c if and only if for all sequences (x_n) in I such that \lim (x_n) = c we have
\left(\frac{f(x_n)-f(c)}{x_n - c}\right)
converges, given x_n \not = c for all n. Furthermore, if f is differentiable at c, then the above sequence of quotients will converge to f^\prime(c).
Let f \colon \mathbb{R} \to \mathbb{R} where f(x) = |x|.
Claim: The function f is non-differentiable at x = 0.
We claim there exists some c \in \mathbb{R} such that f is not differentiabile at c. Using our sequence criterion, we consider the sequence (x_n) = (-1)^n/n.
We have that x_n \to 0 and \forall n,x_n \not = 0. If we assume that this sequence converges, then any subsequence must also converge and converge to the same value. However, when n is even
\frac{f(x_n) - f(0)}{x_n - 0} = \frac{1/n - 0}{1/n -0} = 1
and when n is odd,
\frac{f(x_n) - f(0)}{x_n - 0} = \frac{1/n - 0}{-1/n -0} = -1.
Our subsequences converge to different values, so the original sequence does not converge. Thus f is not differentiable at 0.
The above is an example of a function that is continuous everywhere, but not differentiable everywhere. Below is another example of this phenomonea
Let f \colon \mathbb{R} \to \mathbb{R} where
f(x)=\begin{cases} x\sin(1/x), & x \not = 0\\ 0, & x = 0 \end{cases}
Claim: The function f is non-differentiable at x = 0.
For x \not = 0, we have that
\frac{f(x) - f(0)}{x - 0} = \frac{x\sin(1/x) - 0}{x -0} = \sin(1/x)
But \lim_{x\to 0}\sin(1/x) does not exists. Thus \lim_{x\to 0}\frac{f(x) - f(0)}{x - 0}
does not exists, and f is not differentiable at x = 0.
In the above example, we have a function that is continuous at x = 0, but not differentiable. We can conclude continuity does not imply differentiability.
If f \colon I \to \mathbb{R} is differentiable at c, then f is continuous at c.
We want to show that \lim_{x\to c} f(x) = f(c). We do this by adding zero in a clever way. Note that
f(x) =\left(\frac{f(x)-f(c)}{x-c}\right)(x-c) + f(c)
When x \not = c. Since the derivative f^\prime(c) exists and is finite, we know that the \lim_{x\to c} (f(x)-f(c))/x-c exists and is a real number. Taking the limit of the above equation
\lim_{x \to c}f(x) = \lim_{x\to c}\left(\frac{f(x)-f(c)}{x-c}\right)\lim_{x\to c}(x-c) + \lim_{x \to c}f(c)
Which gives
\lim_{x \to c}f(x) = 0 + f(c) = f(c).
Hence f is continuous at c.
Let f,g \colon I \to \mathbb{R} be differentiable at c \in I. Then
(a) (kf)^\prime(c) = kf^\prime(c), for all k \in \mathbb{R}
(b) (f + g)^\prime(c) = f^\prime(c) + g^\prime(c).
(c) (fg)^\prime(c) = f(c)g^\prime(c) + f^\prime(c)g(c).
(d) (f/g)^\prime(c) = (f(c)g^\prime(c) - f^\prime(c)g(c))/g^2(c), provided g(c) \not = 0
(c) For x \in I, x \not = c, we have \frac{f(x)g(x) - f(c)g(c)}{x-c} = f(x)\left(\frac{g(x) - g(c)}{x-c}\right) + g(c)\left(\frac{f(x)-f(c)}{x-c}\right)
We want to evaluate the limit of the right hand side as x approaches c. Using our limit rules we have
\lim_{x\to c} (fg)^\prime(x) = \lim_{x \to c}f(x)\left(\frac{g(x) - g(c)}{x-c}\right) + \lim_{x \to c} g(c)\left(\frac{f(x)-f(c)}{x-c}\right)
Because f is differentiable at c, f is continuous at c, thus
\lim_{x \to c}f(x)\left(\frac{g(x) - g(c)}{x-c}\right) = f(c)g^\prime(c)
And similarly,
\lim_{x \to c} g(c)\left(\frac{f(x)-f(c)}{x-c}\right) = f^\prime(c)g(c)
Thus
\lim_{x\to c} (fg)^\prime(x) = f(c)g^\prime(c) + f^\prime(c)g(c).
(d)
Let I,J be intervals in \mathbb{R} and f \colon I \to \mathbb{R}, g \colon J \to \mathbb{R}, where f(I)\subseteq J and c \in I. If f is differentiable at c and g is differentiable at f(c), then the composite function g \circ f is differentiable at c and
(g \circ f)^\prime(c) = g^\prime(f(c))f^\prime(c).
Since g is differentiable at f(c), we have
\lim_{y \to f(c)} \frac{g(y)-g(f(c))}{y - f(c)} = g^\prime(f(c)).
Define h \colon J \to \mathbb{R}, where
h(y) = \begin{cases} \frac{g(y)-g(f(c))}{y-f(c)} & \text{if } y \not = f(c)\\ g^\prime(f(c)) & \text{if } y = f(c) \end{cases}
By the first statement, we know that h is continuous at f(c). Since f is differentiable at c, f is continuous at c, and thus
\lim_{x\to c} (h \circ f)(x) = h(f(c)) = g^\prime(f(c)).
Skipped some steps
g(f(c)) - g(f(c)) = h(f(x))(f(x)-f(c))