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Math 554 – Differentiation
Handout #8 – 3/25/
Defn. A function f is said to be differentiable at x 0 if
hlim→ 0
f (x 0 + h) − f (x 0 ) h
exists. In this case the limit is called the derivative of f at x 0 and is denoted f ′(x 0 ).
Note. 1. This definition is equivalent to the requirement that the following limit exist: xlim→x 0
f (x) − f (x 0 ) x − x 0
= f ′(x 0 ).
- This, in turn, is equivalent to the following statement about how fast f (x) converges to f (x 0 ) as x → x 0 : there exists a function η such that limx→x 0 η(x) = 0 and
(∗) f (x) − f (x 0 ) = (x − x 0 ) (f ′(x 0 ) + η(x)).
Examples: 1. If f (x) := x^2 , then f ′(x) = 2x.
- If g(x) := |x|, then g′(0) does not exist.
- If h(x) := x|x|, then h′(x) exists and equals 2|x|.
Theorem. If f is differentiable at x 0 , then f is continuous at x 0. Proof. Use (*) and let x → x 0. 2
Theorem. (Basic rules of differentiation: sums, products, quotients) Suppose that f and g are differentiable at x 0 , then
- (f + g)′(x 0 ) = f ′(x 0 ) + g′(x 0 ).
- (f g)′(x 0 ) = f ′(x 0 )g(x 0 ) + f (x 0 )g′(x 0 ).
- (f /g)′(x 0 ) = (g(x 0 )f ′(x 0 ) − f (x 0 )g′(x 0 )) /g(x 0 )^2 , if g(x 0 ) 6 = 0.
Theorem. (Chain rule) If f is differentiable at x 0 and g is differentiable at y 0 := f (x 0 ), then h := g ◦ f is differentiable at x 0 and
h′(x 0 ) = g′(f (x 0 )) f ′^ (x 0 )
Proof. Use (*) for f at x 0 and for g at y 0 := f (x 0 ):
h(x) − h(x 0 ) x − x 0
g(y) − g(y 0 ) x − x 0
y − y 0 x − x 0
(g′(y 0 ) + η 2 (y))
f (x) − f (x 0 ) x − x 0
(g′(y 0 ) + η 2 (y))
= (f ′(x 0 ) + η 1 (x))(g′(y 0 ) + η 2 (y))
where y := f (x). The proof is completed by using this equation, letting xn → x 0 , and noticing that yn → y 0 where yn := f (xn). 2
Theorem. (Rolle’s Theorem) Suppose that φ is differentiable on (a, b), is contin- uous on [a, b], and vanishes at the endpoints, then there exists x 0 strictly between a and b such that φ′(x 0 ) = 0. Proof. If φ is constant, then any point can be selected for x 0. Otherwise, we may assume WLOG that φ has positive values. By the Extreme Value Theorem, let x 0 be such that φ(x) ≤ φ(x 0 ) for all a ≤ x ≤ b. First, let xn ↓ x 0 , then since x 0 gives a max, we have
0 ≥ φ(xn) − φ(x 0 ) xn − x 0
→ φ′(x 0 )
and so, by the Squeeze Theorem, φ′(x 0 ) ≤ 0. Similarly, φ′(x 0 ) ≥ 0. 2
Note. Within the proof we actually established the critical point procedure of calculus: local max and min can only occur at critical points.
Corollary. (Mean Value Theorem) Suppose that f is differentiable on (a, b) and is continuous on [a, b], then there exists x 0 strictly between a and b such that
f ′(x 0 ) = f (b) − f (a) b − a
Proof. Let
φ(x) := f (x) −
[ (^) f (b)−f (a) b−a (x^ −^ a) +^ f^ (a)
]
and apply Rolle’s theorem. 2
Defn. F is called an anti-derivative of f if F is differentiable and F ′(x) = f (x)
Corollary. If both F and G are anti-derivatives of f , then they differ by a constant, i.e. there exists a constant c such that F (x) − G(x) = c, for all x ∈ dom(f ).
