lacunary - Mathnotes

Differentiation

For f:(a,b)→R,a,b∈R, we say the derivative of f at x, denoted f′(x) is given by

f′(x)=limy→xf(y)−f(x)y−x.

If this limit exists, the function is said to be differentiable at x. If the limit exists for all x∈(a,b), the function is said to be differentiable on (a,b).

Note that this is just the rise-over-run formula for the slope between two points taken to the limit of the two points being infinitesimally near each other.

Theorem: If f is differentiable at x, then it is continuous at x.

Proof: Suppose that f is differentiable at a. We want to show that

limx−>af(x)=f(a).

Now,

limx→a(f(x)−f(a))=limx→a(f(x)−f(a)(x−a)(x−a))…=limx→a(f(x)−f(a)(x−a))⋅limx→a(x−a)…=f′(a)⋅0limx→af(x)−f(a)=0limx→af(x)=f(a).◻

Derivative Rules

Let c be a real number, and f and g be functions defined on an open interval and differentiable at a point x in that interval.

Linearity: (f+cg)′=f′(x)+cg′(x).

Proof:

(f+cg)′(x)=limy→x(f+cg)(y)−(f+cg)(x)y−x=limy→xf(y)+(cg)(y)−f(x)−(cg)(x)y−x=limy→xf(y)−f(x)y−x+limy→xcg(y)−cg(x)y−x=f′(x)+cg′(x). ◻

Product: (fg)′(x)=f′(x)g(x)+f(x)g′(x).

Proof: The trick here is to subtract and add f(y)g(x) to the numerator on the second line. This allows factoring, and then we just use limit laws.

(fg)′(x)=limy→x(fg)(y)−(fg)(x)y−x=limy→xf(y)g(y)−f(y)g(x)+f(y)g(x)−f(x)g(x)y−x=limy→x(g(y)−g(x))f(y)+(f(y)−f(x))g(x)y−x=limy→xg(x)f(y)−f(x)y−x+limy→xf(y)g(y)−g(x)y−x=limy→xg(x)⋅limy→xf(y)−f(x)y−x+limy→xf(y)⋅limy→xg(y)−g(x)y−x=f′(x)g(x)+f(x)g′(x). ◻

Quotient: (f/g)′(x)=f′(x)g(x)−f(x)g′(x)[g(x)]2.

Proof: Since division is multiplication by a reciprocal, we use the same trick as for the product rule.

(f/g)′(x)=limy→x(f/g)y−(f/g)xy−x=limy→xf(y)g(y)−f(x)g(x)y−x=limy→xf(y)g(x)−f(x)g(y)g(y)g(x)y−x=limy→xf(y)g(x)−f(x)g(x)+f(x)g(x)−f(x)g(y)g(y)g(x)y−x=limy→xf(y)g(x)−f(x)g(x)g(y)g(x)y−x−limy→x(f(x)g(y)−f(x)g(x)g(y)g(x)y−x)=limy→x(f(y)−f(x)y−xg(x)g(y)g(x))−limy→x(g(y)−g(x)y−xf(x)g(y)g(x))=f′(x)g(x)−g′(x)f(x)[g(x)]2. ◻

Let h(x)=f(g(x)).

Chain: h′(x)=f′(g(x))⋅g′(x).

Local Extrema

The real valued function on the real line f is said to have a local maximum at c∈R if there exists a δ>0 such that f(x)≤f(c) for all x∈(c−δ,c+δ).

The real valued function on the real line f is said to have a local minimum at c∈R if there exists a δ>0 such that f(x)≥f(c) for all x∈(c−δ,c+δ).

Derivatives at Local Extrema Theorem (Fermat's Theorem): Suppose f is defined on an open interval, that c is a number in that interval, and that f has a local maximum or minimum at c, and that f′(c) exists. Then f′(c)=0.

Other Theorems

Rolle's Theorem: Suppose f is continuous on [a,b] and differentiable on (a,b) and that f(a)=f(b). Then there exists a point c in (a,b) such that f′(c)=0.

Simply put, if a differentiable function over an open interval (a,b) has the same value at its boundary points, then at some point in the interval, the function is flat (has a zero derivative).

Theorem: Mean Value Theorem \@{mean-value-theorem}

Suppose f is continuous on [a,b] and differentiable on (a,b). Then there exists a point c in (a,b) such that:

f′(c)=f(b)−f(a)b−a.

This means that for a function continuous on an interval, and differentiable on that interval except maybe at its endpoints, there is a point on the interval where the derivative of the function at that point equals the slope of the function between its endpoints.

Referenced by (2 direct)

Direct references:

Theorem: L'Hospital \@{lhospital}

Suppose there is a δ>0 such that f and g are differentiable on (c−δ,c+δ) with g′(x)≠0 on this interval. Suppose also that

limx→cf′(x)g′(x)=L.

Then if limx→cf(x)=limx→cg(x)=0, limx→cf(x)=limx→cg(x)=∞, or limx→cf(x)=limx→cg(x)=−∞, we have that

limx→cf(x)g(x)=L.

The endpoint c can also be ±∞.

Theorem: Real Valued Continuously Differentiable implies Lipschitz Continuous \@{real-valued-continuously-differentiable-implies-lipschitz-continuous}

Suppose f:[a,b]→R is continuously differentiable. Then f is Lipschitz continuous with a Lipschitz constant

L=supx∈[a,b]|f′(x)|.

Proof \@{proof-of-real-valued-continuously-differentiable-implies-lipschitz-continuous}

Because [a,b] is compact and f′ is continuous, by the Extreme Value Theorem we have that f′ attains a minimum and a maximum on [a,b], which gives us L (actually a stronger condition: L=maxx∈[a,b]|f′(x)|=supx∈[a,b]|f′(x)|). By the Mean Value Theorem, for any (x1,x2)⊂[a,b], there exists some c such that f(x2)−f(x1)=f′(c)(x2−x1).The EVT implies that |f′(c)|≤L, so|f(x2)−f(x1)|≤L|x2−x1|, which is what we wanted to show.

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Theorem: Mean Value Inequality for Vector Valued Functions \@{mean-value-inequality-for-vector-valued-functions}

Suppose f:[a,b]→Rd is continuous on [a,b] and differentiable on (a,b). Then there exists x∈(a,b) such that ‖f(b)−f(a)‖≤(b−a)‖f′(x)‖.

Proof \@{proof-of-mean-value-inequality-for-vector-valued-functions}

Let z=f(b)−f(a), and define φ(t)=⟨z,f(t)⟩,(a≤t≤b). If z=0, then the result is trivial and we're done, so assume z≠0. Then φ is a real valued continuous function on [a,b] and is differentiable on (a,b). By the Mean Value Theorem, there is some x∈(a,b) such that φ(b)−φ(a)=(b−a)φ′(x)=(b−a)⟨z,f′(x)⟩. On the other hand, φ(b)−φ(a)=⟨z,f(b)⟩−⟨z,f(a)⟩=⟨z,f(b)−f(a)⟩=⟨z,z⟩=‖z‖2. Now, by the Cauchy-Schwarz Inequality, ‖z‖2=(b−a)|⟨z,f′(x)⟩|≤(b−a)‖z‖‖f′(x)‖. Therefore, ‖z‖≤(b−a)‖f′(x)‖, which is what we wanted to show.

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Referenced by (1 direct)

Direct references:

Theorem: Vector Valued Continuously Differentiable implies Lipschitz Continuous \@{vector-valued-continuously-differentiable-implies-lipschitz-continuous}

Suppose f:[a,b]→Rd is continuously differentiable. Then f is Lipschitz continuous with a Lipschitz constant

L≤supx∈[a,b]‖f′(x)‖

where f′ is the Jacobian matrix of f and ‖⋅‖ denotes the matrix norm induced by the @euclidean inner product.

Proof \@{proof-of-vector-valued-continuously-differentiable-implies-lipschitz-continuous}

Because [a,b] is compact and ‖f′‖ is continuous on it, by the Extreme Value Theorem we have that ‖f′‖ attains a minimum and a maximum on [a,b]; then let L=maxx∈[a,b]‖f′(x)‖=supx∈[a,b]‖f′(x)‖. Now, by Mean Value Inequality for Vector Valued Functions, for any (x1,x2)⊂[a,b], there exists some c∈(a,b) such that ‖f(x2)−f(x1)‖≤(x2−x1)‖f′(c)‖. But ‖f′(c)‖≤L, so ‖f(x2)−f(x1)‖≤L|x2−x1|, which is what we wanted to show.

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