Higher Order Derivatives

Higher-Order Derivatives: Rigorous Definition and Notation

If a function \( y = f(x) \) is differentiable, we can compute its first derivative \( f'(x) \) or \( \frac{dy}{dx} \), which represents the rate of change of \( y \) with respect to \( x \). If this derivative itself is differentiable, we can differentiate it again to obtain the second derivative, and this process can be continued to define derivatives of higher order.

Second-Order Derivative

The second derivative is defined as the derivative of the first derivative:

\[ \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d^2y}{dx^2} \]

or equivalently, in function notation:

\[ \frac{d}{dx} \left( f'(x) \right) = f''(x) \]

This represents the rate of change of the first derivative, which provides information about the curvature or concavity of the function (we will learn later the correct interpretation).

Third-Order Derivative

If the second derivative exists and is differentiable, we can differentiate it again:

\[ \frac{d}{dx} \left( \frac{d^2y}{dx^2} \right) = \frac{d^3y}{dx^3} \]

or in function notation:

\[ \frac{d}{dx} \left( f''(x) \right) = f'''(x) \]

\( n \)th-Order Derivative

Proceeding inductively, if the \((n-1)\)th derivative exists and is differentiable, then the \( n \)th-order derivative is defined as:

\[ \frac{d}{dx} \left( \frac{d^{n-1}y}{dx^{n-1}} \right) = \frac{d^n y}{dx^n} \]

or, equivalently,

\[ \frac{d}{dx} \left( f^{(n-1)}(x) \right) = f^{(n)}(x) \]

where \( f^{(n)}(x) \) represents the \( n \)th derivative of \( f(x) \).

Thus, the process of differentiation can be iterated indefinitely, provided that the required derivatives exist. The notation follows a structured pattern:

• For small values of \( n \):

  \[ f'(x), \quad f''(x), \quad f'''(x) \]

• For \( n \geq 4 \), parentheses are used:

  \[ f^{(4)}(x), \quad f^{(5)}(x), \quad \dots, \quad f^{(n)}(x) \]

• In Leibniz notation, the general form is:

  \[ \frac{d^n y}{dx^n} \]

which emphasizes that the differentiation operator \( \frac{d}{dx} \) is applied \( n \) times.

nth order derivatives of some common functions

Let us now calculate the \( n \)th-order derivatives of some common functions by differentiating repeatedly and then generalizing the pattern using induction.

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1. Power Function: \( f(x) = x^m \)

We begin with a power function \( f(x) = x^m \), where \( m \) is any real number. The first derivative is:

\[ f'(x) = m x^{m-1}. \]

Differentiating again,

\[ f''(x) = m (m-1) x^{m-2}. \]

Continuing this process, we observe the pattern:

\[ f^{(n)}(x) = m (m-1) (m-2) \dots (m-n+1) x^{m-n}. \]

Using factorial notation, we write:

\[ f^{(n)}(x) = \frac{m!}{(m-n)!} x^{m-n}, \quad \text{for } m \geq n. \]

For \( n > m \), the derivative vanishes:

\[ f^{(n)}(x) = 0, \quad \text{for } n > m. \]

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2. Exponential Function: \( f(x) = e^x \)

The function \( e^x \) is unique in that differentiating it any number of times gives the same function:

\[ f'(x) = e^x, \quad f''(x) = e^x, \quad f'''(x) = e^x, \quad \dots \]

Thus, for any \( n \geq 1 \),

\[ f^{(n)}(x) = e^x. \]

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3. General Exponential Function: \( f(x) = a^x \)

For a constant base \( a > 0 \), the first derivative is:

\[ f'(x) = (\ln a) a^x. \]

Differentiating again,

\[ f''(x) = (\ln a)^2 a^x. \]

By induction, the \( n \)th derivative follows as:

\[ f^{(n)}(x) = (\ln a)^n a^x. \]

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4. Reciprocal Function: \( f(x) = \frac{1}{x} \)

Rewriting as \( f(x) = x^{-1} \), the first derivative is:

\[ f'(x) = -x^{-2} = -\frac{1}{x^2}. \]

Differentiating again,

\[ f''(x) = 2 x^{-3} = \frac{2}{x^3}. \]

Proceeding inductively, we obtain:

\[ f^{(n)}(x) = (-1)^n \frac{n!}{x^{n+1}}, \quad n \geq 1. \]

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5. Reciprocal of a Linear Function: \( f(x) = \frac{1}{ax+b} \)

Using the chain rule, the first derivative is:

\[ f'(x) = -\frac{a}{(ax+b)^2}. \]

Differentiating again,

\[ f''(x) = \frac{2a^2}{(ax+b)^3}. \]

Observing the pattern, we generalize:

\[ f^{(n)}(x) = (-1)^n \frac{n! a^n}{(ax+b)^{n+1}}, \quad n \geq 1. \]

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6. Sine Function: \( f(x) = \sin x \)

The derivatives of \( \sin x \) follow a cyclic pattern:

\[ f'(x) = \cos x, \quad f''(x) = -\sin x, \quad f'''(x) = -\cos x, \quad f^{(4)}(x) = \sin x. \]

Extending this to the \( n \)th derivative, we observe:

\[ f^{(n)}(x) = \sin \left( x + n \frac{\pi}{2} \right). \]

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7. Cosine Function: \( f(x) = \cos x \)

Similar to the sine function, the derivatives of \( \cos x \) repeat every four steps:

\[ f'(x) = -\sin x, \quad f''(x) = -\cos x, \quad f'''(x) = \sin x, \quad f^{(4)}(x) = \cos x. \]

Thus, the general formula for the \( n \)th derivative is:

\[ f^{(n)}(x) = \cos \left( x + n \frac{\pi}{2} \right). \]

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8. Product Function: \( f(x) = x e^x \)

We now compute the \( n \)th derivative of \( f(x) = x e^x \). Using the product rule:

\[ f'(x) = (x+1)e^x. \]

Differentiating again,

\[ f''(x) = (x+2)e^x. \]

Repeating this process, we obtain the general form:

\[ f^{(n)}(x) = (x+n) e^x. \]

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Summary of Results

Function	\( n \)th Derivative
\( x^m \)	\( \frac{m!}{(m-n)!} x^{m-n}, \quad n \leq m \); 0 otherwise
\( e^x \)	\( e^x \)
\( a^x \)	\( (\ln a)^n a^x \)
\( \frac{1}{x} \)	\( (-1)^n \frac{n!}{x^{n+1}} \)
\( \frac{1}{ax+b} \)	\( (-1)^n \frac{n! a^n}{(ax+b)^{n+1}} \)
\( \sin x \)	\( \sin \left( x + n \frac{\pi}{2} \right) \)
\( \cos x \)	\( \cos \left( x + n \frac{\pi}{2} \right) \)
\( x e^x \)	\( (x+n) e^x \)

These formulas illustrate how different types of functions exhibit distinct patterns in their higher-order derivatives. Understanding these patterns allows for efficient computation of derivatives in various mathematical contexts.

Leibniz Rule for product of two fucntions

Let us compute the derivatives of a product of two differentiable functions \( f(x) \) and \( g(x) \) systematically, starting with the first few derivatives and then generalizing to the \( n \)th-order case.

First Derivative (Standard Product Rule)

Applying the product rule:

\[ \frac{d}{dx} [f(x) g(x)] = f'(x) g(x) + f(x) g'(x). \]

This formula states that the derivative of a product is the sum of two terms: one where we differentiate \( f(x) \) while keeping \( g(x) \) fixed, and another where we differentiate \( g(x) \) while keeping \( f(x) \) fixed.

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Second Derivative

Differentiating both sides again:

\[ \frac{d^2}{dx^2} [f(x) g(x)] = \frac{d}{dx} \left[ f'(x) g(x) + f(x) g'(x) \right]. \]

Applying the product rule to each term:

\[ = f''(x) g(x) + f'(x) g'(x) + f'(x) g'(x) + f(x) g''(x). \]

Rearranging,

\[ \frac{d^2}{dx^2} [f(x) g(x)] = f''(x) g(x) + 2 f'(x) g'(x) + f(x) g''(x). \]

Notice that the coefficient 2 appears in front of the middle term.

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Third Derivative

Differentiating again:

\[ \frac{d^3}{dx^3} [f(x) g(x)] = \frac{d}{dx} \left[ f''(x) g(x) + 2 f'(x) g'(x) + f(x) g''(x) \right]. \]

Applying the product rule to each term:

\[ = f'''(x) g(x) + f''(x) g'(x) + 2 f''(x) g'(x) + 2 f'(x) g''(x) + f'(x) g''(x) + f(x) g'''(x). \]

Rearranging,

\[ \frac{d^3}{dx^3} [f(x) g(x)] = f'''(x) g(x) + 3 f''(x) g'(x) + 3 f'(x) g''(x) + f(x) g'''(x). \]

The coefficients 1, 3, 3, 1 are exactly the third row of Pascal’s triangle (binomial coefficients).

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Pattern and Generalization

Observing the derivatives, we notice the general structure:

\[ \frac{d^n}{dx^n} [f(x) g(x)] = \sum_{k=0}^{n} \binom{n}{k} f^{(k)}(x) g^{(n-k)}(x). \]

where \( \binom{n}{k} \) is the binomial coefficient:

\[ \binom{n}{k} = \frac{n!}{k!(n-k)!}. \]

This follows by induction and generalizes the product rule to higher-order derivatives.

Thus, for any order \( n \), the derivative of a product follows the structure of expanding a binomial expression, where each term consists of a combination of derivatives of \( f(x) \) and \( g(x) \) multiplied by the corresponding binomial coefficient. COnsider the following example:

Let \( f(x) = x^2 \) and \( g(x) = e^x \). We compute \( \frac{d^3}{dx^3} (x^2 e^x) \):

Applying Leibniz's rule:

\[ \frac{d^3}{dx^3} (x^2 e^x) = \sum_{k=0}^{3} \binom{3}{k} f^{(k)}(x) g^{(3-k)}(x). \]

Computing individual terms:

• \( f(x) = x^2 \Rightarrow f^{(0)}(x) = x^2, \quad f^{(1)}(x) = 2x, \quad f^{(2)}(x) = 2, \quad f^{(3)}(x) = 0 \).
• \( g(x) = e^x \Rightarrow g^{(n)}(x) = e^x \) for all \( n \).

Substituting:

\[ \frac{d^3}{dx^3} (x^2 e^x) = \binom{3}{0} x^2 e^x + \binom{3}{1} 2x e^x + \binom{3}{2} 2 e^x + \binom{3}{3} 0. \]

Simplifying:

\[ (x^2 + 6x + 6) e^x. \]

Double Derivative of Inverse of a Function

Let \( f \) be an invertible function, and let \( g \) be its inverse, so that \( g(x) = f^{-1}(x) \). We already know that the first derivative of \( g(x) \) is given by

\[ g'(x) = \frac{1}{f'(g(x))}. \]

Differentiating both sides again,

\[ g''(x) = \frac{d}{dx} \left( \frac{1}{f'(g(x))} \right). \]

Applying the chain rule,

\[ g''(x) = -\frac{1}{[f'(g(x))]^2} \cdot \frac{d}{dx} [f'(g(x))]. \]

Since \( f'(g(x)) \) is a composite function, differentiating it gives

\[ \frac{d}{dx} [f'(g(x))] = f''(g(x)) g'(x). \]

Substituting this back,

\[ g''(x) = -\frac{1}{[f'(g(x))]^2} \cdot f''(g(x)) g'(x). \]

Using the previously known result \( g'(x) = \frac{1}{f'(g(x))} \), we substitute \( g'(x) \):

\[ \boxed{g''(x) = -\frac{f''(g(x))}{[f'(g(x))]^3}}. \]

This result expresses the second derivative of an inverse function purely in terms of the derivatives of \( f(x) \), evaluated at \( g(x) \).

Double Derivative of a Parametric Function

Let \( x = f(t) \) and \( y = g(t) \) be a parametrically defined function. We already know that the first derivative of \( y \) with respect to \( x \) is given by

\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{g'(t)}{f'(t)}. \]

Differentiating this again with respect to \( x \),

\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{g'(t)}{f'(t)} \right). \]

Since \( t \) is an implicit function of \( x \), we use the chain rule,

\[ \frac{d^2y}{dx^2} = \frac{d}{dt} \left( \frac{g'(t)}{f'(t)} \right) \frac{dt}{dx}. \]

Applying the quotient rule to differentiate \( \frac{g'(t)}{f'(t)} \),

\[ \frac{d}{dt} \left( \frac{g'(t)}{f'(t)} \right) = \frac{f'(t) g''(t) - g'(t) f''(t)}{[f'(t)]^2}. \]

Since \( \frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{f'(t)} \), substituting this,

\[ \frac{d^2y}{dx^2} = \frac{f'(t) g''(t) - g'(t) f''(t)}{[f'(t)]^2} \cdot \frac{1}{f'(t)}. \]

Simplifying,

\[ \frac{d^2y}{dx^2} = \frac{f'(t) g''(t) - g'(t) f''(t)}{[f'(t)]^3}. \]

This expresses the second derivative of a parametrically defined function in terms of the first and second derivatives of \( f(t) \) and \( g(t) \).