Higher derivatives

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Recall

The derivative of a function at a specific input value describes the instantaneous rate of change. Geometrically, when the derivative exists, it is the slope of the tangent line to the graph of the function at that point. Algebraically, it represents the ratio of the change in the dependent variable to the change in the independent variable as the change approaches zero.

Definition. A function f(x) is differentiable at a point “a” in its domain, if its domain contains an open interval around “a”, and the following limit exists $f'(a) = L = \lim _{h \to 0}{\frac {f(a+h)-f(a)}{h}}$. More formally, for every positive real number ε, there exists a positive real number δ, such that for every h satisfying 0 < |h| < δ, then the following inequality holds |L-$\frac {f(a+h)-f(a)}{h}$|< ε.

Basic important derivatives

1. Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$ (works for any real n).
2. Sum Rule: $\frac{d}{dx}(f(x) + g(x)) = \frac{d}{dx}(f(x)) + \frac{d}{dx}(g(x))$
3. Difference Rule: $\frac{d}{dx}(f(x) - g(x)) = f'(x) - g'(x)$
4. Constant Multiple Rule: $\frac{d}{dx}(c \cdot f(x)) = c \cdot f'(x)$ where c is a constant.
5. Product Rule: $\frac{d}{dx}(f(x) \cdot g(x)) = f'(x)g(x) + f(x)g'(x)$.
6. Quotient Rule: $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$, provided $g(x) \ne 0$.
7. Chain Rule: $\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)$
8. $\frac{d}{dx}(e^x) = e^x, \frac{d}{dx}(\ln(x)) = \frac{1}{x}, \frac{d}{dx}(\sin(x)) = \cos(x), \frac{d}{dx}(\cos(x)) = -\sin(x), \frac{d}{dx}(\tan(x)) = \sec^2(x), \frac{d}{dx}(\arcsin(x)) = \frac{1}{\sqrt{1 - x^2}}, \frac{d}{dx}(\arccos(x)) = -\frac{1}{\sqrt{1 - x^2}}, \frac{d}{dx}(\arctan(x)) = \frac{1}{1 + x^2}.$

Higher Order Derivatives.

Differentiation is not a one-time process. Since the derivative of a function y = f(x) is itself a function, we can differentiate it again, which is generally referred to as the second derivative of f and written as f"(x), f²(x), or even simply f". This differentiation process can be continued to find the third, fourth, and so on. This leads us to the concept of higher-order derivatives.

• The first derivative of f(x), denoted f’(x) or $\frac{dy}{dx}$, measures the rate of change of f with respect to x.
• The second derivative, written f’’(x), $\frac{d^2y}{dx^2}$, or $D^2y$, measures the rate of change of the first derivative.
• This process can be continued: $f^{(3)}(x)=\frac{d^3y}{dx^3}, f^{(4)}(x)=\frac{d^4y}{dx^4}$ and so on. Collectively, all derivatives beyond the first —second, third, fourth, etc.— are called higher-order derivatives.

y" = $\frac{d}{dx}\frac{dy}{dx}=\frac{d}{dx}\frac{d}{dx}y = (\frac{d}{dx})^{2}y = \frac{d^{2}y}{dx^{2}}$, y³ = $\frac{d^{3}y}{dx^{3}} = D^{3}y$ and so on.

The operator $\frac{d}{dx}$ is the differential operator. $\frac{d}{dx}:f(x) \to f′(x)$ It is often treated as an abstract operation or machine that accepts a function (or takes a function as input) and returns another function as output.

Geometric and Physical Interpretation

The first derivative of a function f(x), denoted as f'(x), represents the rate of change of f(x) with respect to x (how fast f(x) is changing at each point). Geometrically, it is the slope of the tangent line to the graph of f(x) at each point x.

For example, if we are studying the motion of an object and we plot the distance traveled against time s(t), the slope of the graph $s'(t)=\frac{ds}{dt}$ represents the speed or velocity of the object (e.g., how fast a car is moving). A steeper slope indicates a higher speed, while a flatter slope indicates a lower speed.

If R(t) is a company’s revenue at time t:

• R′(t)>0 means that revenue is increasing at time t.
• If the rate of change of a company's revenue is increasing (i.e., R′(t) itself is increasing), then the company’s growth is accelerating.

The second derivative, denoted as f’’(x) or $\frac{d^2f}{dx^2}$ represents the rate of change of the first derivative f’(x) and describes how the slope of the tangent line to the graph of f is changing or, in other words, it measures the curvature or concavity of the function..

At points where it curves upwards (shaped like a cup), the slope is increasing and that means the second derivative is positive f’’(x) > 0. On the other hand, at points where it curves downwards (shaped like a frown), the slope is decreasing and that means the second derivative is negative f’’(x) < 0.

Back to our previous example, if we are studying the motion of an object and we plot the distance traveled against time s(t), the second derivate is the object’s acceleration $s''(t) = \frac{d^2s}{dt^2}$, how the velocity changes over time. Therefore, we have positive acceleration whenever the car is speeding up (increasing velocity) and negative acceleration whenever the car is slowing down (decreasing velocity).

A Journey from Madrid to Barcelona

Imagine a car traveling from Madrid to Barcelona:

• At first, the car starts slowly: the graph of s(t) is shallow.
• As it accelerates, the curve steepens —distance increases rapidly.
• Near the end, the car decelerates: the curve flattens again

We could graph this motion (s(t)), letting the vertical axis (y) represents the distance traveled by the vehicle (measured in units such as meters, kilometers, miles, etc.), and the horizontal axis (x) represents the time taken to travel the distance (measured in units such as seconds, minutes, hours, etc.).

Initially the curve is quite shallow (commercial cars are typically slow to start), then for the next seconds, as the car speeds up, the distance traveled in a given time get larger or, in other words, the curve become steeper, and finally, as the car slows towards the end, the curve shallows out again, so the graph $\frac{ds}{dt}$ might look like a bump, increasing up to some maximum, then decreasing back to zero (Figure iii).

The acceleration’s curve is positive for the first half of the journey which indicates our vehicle is speeding up, then negative which indicates the car is slowing down.

Solved exercises

• Polynomials. $y = x^{m}, y' = mx^{m-1}.~ y'' = m(m-1)x^{m-2}.~ y''' =m(m-1)(m-2)x^{m-3}.~ y^{n}=m·(m-1)···(m-n+1)x^{m-n}.$

Induction. The formula is valid for n = 1. Assuming that it is true for n, let’s differentiate again.

$y^{n+1}=m·(m-1)···(m-n+1)(m-n)x^{m-n-1} = m·(m-1)···(m-n+1)(m-n)x^{m-(n+1)}$, that is, the derivative of (n+1)-th order is expressed by the same formula as the n-th order derivative (n is just replaced by n +1 ).

Example 1: Let $f(x) = x^4 - 3x^2 + 2$. Compute higher-order derivatives:

1. First derivative: $f'(x) = 4x^3 - 6x$.
2. Second derivative: $f''(x) = 12x^2 - 6$.

   f’’(x) > 0 when $12x^2 - 6 > 0 \iff |x| > \sqrt{\tfrac{1}{2}}$: graph is concave up there. f’’(x) < 0 when $|x| < \sqrt{\tfrac{1}{2}}$: graph is concave down there. Points where f’’(x) = 0 are candidates for inflection points.

3. Third derivative: $f^{(3)}(x) = 24x$.
4. Fourth derivative: $f^{(4)}(x) = 24$.
5. Fifth derivative and higher: $f^{(n)}(x) = 0 \quad \text{for all } n \ge 5$

Example 2: $f(x) = x^5 - 3x^3 + 2x^2 - 7x + 4$

• Exponential Function. $f(x) = e^x$. $f^{(n)}(x), \forall n \ge 1$, all derivatives of $e^x$ are $e^x$.

In particular, for $f(x) = e^{kx}, \boxed{f^{(n)}(x) = k^n e^{kx}}$

Exercise: Find the 100th derivate of y = e^-2x.

y’ = -2e^-2x, y’’ = (-2)²e^-2x, y’’’ = (-2)³e^-2x […] y⁽¹⁰⁰⁾ =[k = -2, $f(x) = e^{kx}, \boxed{f^{(n)}(x) = k^n e^{kx}}$] (-2)¹⁰⁰e^-2x = 2¹⁰⁰e^-2x.

• Trigonometric Functions. $f(x) = \sin(x)$

We observe that the derivatives of sin(x) repeat every four derivatives due to the nature of the sine function:

$f^{(n)}(x) = \begin{cases} \sin(x), & n \equiv 0 \pmod{4} \\ \cos(x), & n \equiv 1 \pmod{4} \\ -\sin(x), & n \equiv 2 \pmod{4} \\ -\cos(x), & n \equiv 3 \pmod{4} \end{cases}$

Example 1: n = 50 = 4·12 + 2, $y^{(4n+2)} = y^{(4·12+2)} = -sin(x),$ n = 51 = 4·12 + 3, $y^{(4n+3)} = y^{(4·12+3)} = -cos(x).$

Example 2: Find the 99th and 100th derivate of y = sin(3x), y’ = 3cos(3x), y" = -3²sin(x), y³ = -3³cos(3x), y⁴ = 3⁴sin(3x) [···] We observe that the pattern repeats every four derivatives due to the nature of the sine function. Applying a similar argument to that used in the previous example, y⁹⁹ = y^{4·24 +3} =[y³ = -3³cos(3x)] -3⁹⁹cos(3x), y¹⁰⁰ = y^5·24 =[y⁴ = 3⁴sin(3x)] = 3¹⁰⁰sin(3x).

• Natural Logarithm. $f(x) = \ln(x)$

Pattern Recognition: (i) signs alternate starting with positive for n = 1 (+,−,+,−,… $\to (-1)^{n-1}$); (ii) coefficients are factorials of n -1; (iii) denominator $x^n$ (exponent increases by 1 with each derivative).

• y = $\frac{7}{x^2} = 7x^{-2}$. Apply the power rule: $\frac{d}{dx}x^n = nx^{n-1}$ and we get: y’ = $7·(-2)x^{-3} = \frac{(-1)^1·7·2·1}{x^3} = \frac{-14}{x^3}, y''=-14·(-3)x^{-4} = \frac{42}{x^4} = \frac{(-1)^2·7·3·2·1}{x^4}, y''' = 42·(-4)x^{-5} = \frac{-168}{x^{5}} = \frac{(-1)^3·7·4·3·2·1}{x^4}$, and so on. As you can see, the pattern continues with: (i) alternating signs with each derivate ($(-1)^n$); (ii) coefficients start at 7, then multiples by 2, then 3, … ($7 \cdot (n+1)!$); and (iii) increasing powers of x in the denominator ($x^{n+1}$). The nth derivative can be expressed by the general formula: $\boxed{f^{(n)}(x) = (-1)^n \frac{7·(n+1)!}{x^{n+2}}}$

• y = $\frac{1}{x-1} = (x-1)^{-1}$

y’ = $(-1)(x-1)^{-2}, y''=2·(x-1)^{-3}, y''' = 2·(-3)(x-1)^{-4} = (-6)(x-1)^{-4}, y'''' = (-6)·(-4)(x-1)^{-4}=24·(x-1)^{-4}$, and so on, hence: (i) signs alternate with each derivate ($(-1)^n$); (ii) coefficients grow factorially because each derivative multiplies by the next integer 1!, 2!, 3!,… (n!); and (iii) denominator equals $(x-1)^{n+1}$ (since the original exponent is −1, and each derivative reduces it by 1). General formula: $\boxed{f^{(n)}(x) = \frac{(-1)^n·n!}{(x-1)^{(n+1)}}}$

• y = $xe^x$

Apply the product rule and we get: y’ = $e^x+xe^x =(x+1)e^x, y'' =e^x +(x+1)e^x = (x+2)e^x, y''' = e^x + (x+2)e^x = (x+3)e^x$. Pattern Recognition: (i) each derivative retains the form (x + constant)$e^n$; (ii) the constant term increases by 1 with each derivative (n); (iii) the exponential term remains unchanged, so the general formula is $\boxed{y^{(n)} = (x+n)e^x}.$

Applications of Higher Derivatives

• Concavity and Inflection Points. f’: rate of change of f; slope of the tangent line; tells you whether the function is increasing or decreasing. f’’: rate of change of f′(x); describes curvature/concavity. An inflection point occurs where concavity changes:

Example: $f(x) = x^3$. $f''(x) = 6x, f''(0) = 0$. $f''(x) < 0$ for $x < 0$, $f''(x) > 0$ for $x > 0$. Inflection point at $x = 0$ ✓ This is the archetypal example where the graph transitions smoothly from bending down to bending up.

• Second Derivative Test for Extrema. At a critical point where $f'(c) = 0$, the second derivative tells us the shape of the graph there (concavity determines whether the tangent line sits above or below the graph near the critical point).

• Taylor Series. An application of higher-order derivatives is in the Taylor series expansions, which are used to approximate functions with polynomials. The Taylor series of a function at a point x=a is: $f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + ··· + \frac{f^{(n)}(a)}{n!}(x-a)^n + ··· = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$.

Taylor series are a way to represent a function as an infinite sum of terms that are expressed in terms of the function’s derivatives at a single point. For most common functions, the function and the sum of its Taylor series are (almost) equal near this point, e.g., e^x =[e’(x) = e(x), e’’(x) = e(x), and so on] $\sum_{n=0}^∞ \frac{f^{(n)}(0)}{n!}x^n = \sum_{n=0}^∞ \frac{1}{n!}x^n$; sin(x) =[sin’(x) = cos(x), sin’’(x) = -sin(x), sin’’’(x) = -cos(x), sin⁴(x) = sin(x) and the pattern repeat, at x = 0, we get sin(0) = 0, sin’(0) = 1, 0, -1, 0] $\sum_{n=0}^∞ \frac{f^{(n)}(0)}{n!}x^n = x - \frac{x^3}{3!} + \frac{x^5}{5!}-\frac{x^7}{7!}+··· = \sum_{n=0}^∞ \frac{(-1)^n·x^{2n+1}}{(2n+1)!}$

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