Miscellaneuous

Derivative of ln|x|

The function \( \ln |x| \) is defined for all \( x \in \mathbb{R} \setminus \{0\} \) and can be expressed as:

\[ \ln |x| = \begin{cases} \ln x, & x > 0 \\ \ln(-x), & x < 0 \end{cases} \]

Differentiating each case separately:

For \( x > 0 \), we have

\[ \frac{d}{dx} (\ln x) = \frac{1}{x} \]

For \( x < 0 \), we have

\[ \frac{d}{dx} (\ln (-x)) = \frac{1}{-x} \cdot (-1) = \frac{1}{x} \]

Thus, for all \( x \neq 0 \):

\[ \frac{d}{dx} (\ln |x|) = \frac{1}{x}, \quad x \in \mathbb{R} \setminus \{0\} \]

Differentiating Determinants

Let us consider the determinant of a \( 2 \times 2 \) matrix whose entries are functions of \( x \):

\[ D(x) = \begin{vmatrix} a(x) & b(x) \\ c(x) & d(x) \end{vmatrix} = a(x)d(x) - b(x)c(x). \]

Differentiating both sides with respect to \( x \),

\[ D'(x) = a'(x) d(x) + a(x) d'(x) - b'(x) c(x) - b(x) c'(x). \]

Now, grouping the first and fourth terms together and the second and third terms together:

\[ D'(x) = \left( a'(x) d(x) - b(x) c'(x) \right) + \left( a(x) d'(x) - b'(x) c(x) \right). \]

Observing each group separately, we recognize them as determinants:

\[ D'(x) = \begin{vmatrix} a'(x) & b(x) \\ c'(x) & d(x) \end{vmatrix} + \begin{vmatrix} a(x) & b'(x) \\ c(x) & d'(x) \end{vmatrix}. \]

This shows that the derivative of the determinant can be computed by differentiating column-wise and summing the resulting determinants.

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Alternatively, we can group the first and third terms together and the second and fourth terms together:

\[ D'(x) = \left( a'(x) d(x) - b'(x) c(x) \right) - \left( a(x) d'(x) - b(x) c'(x) \right). \]

Rewriting these as determinants,

\[ D'(x) = \begin{vmatrix} a'(x) & b'(x) \\ c(x) & d(x) \end{vmatrix} + \begin{vmatrix} a(x) & b(x) \\ c'(x) & d'(x) \end{vmatrix}. \]

This expresses the derivative as the sum of two determinants obtained by differentiating row-wise.

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Thus, we observe the foloowing rule:

• The derivative of a determinant can be computed by summing determinants obtained by differentiating column-wise.
• Equivalently, it can be computed by summing determinants obtained by differentiating row-wise.

Extending this to an \( n \times n \) determinant, we obtain the general formula:

\[ \frac{d}{dx} \det A(x) = \sum_{j=1}^{n} \det A_j(x), \]

where \( A_j(x) \) is the matrix obtained by differentiating the \( j \)th row while keeping the others unchanged. The same formula holds when differentiating column-wise instead of row-wise.

For example:

Consider the \( 2 \times 2 \) matrix:

\[ A(x) = \begin{bmatrix} x^2 & e^x \\ \sin x & \ln x \end{bmatrix}. \]

The determinant of \( A(x) \) is:

\[ D(x) = \begin{vmatrix} x^2 & e^x \\ \sin x & \ln x \end{vmatrix}. \]

To differentiate \( D(x) \), we first differentiate column-wise:

\[ D'(x) = \begin{vmatrix} 2x & e^x \\ \cos x & \ln x \end{vmatrix} + \begin{vmatrix} x^2 & e^x \\ \sin x & \frac{1}{x} \end{vmatrix}. \]

Now, differentiating row-wise:

\[ D'(x) = \begin{vmatrix} 2x & e^x \\ \sin x & \frac{1}{x} \end{vmatrix} + \begin{vmatrix} x^2 & e^x \\ \cos x & \ln x \end{vmatrix}. \]

Both formulations give the same result, verifying the determinant differentiation rule.

Summation of the Series Using Differentiation

Consider the infinite series

\[ 1 + 2x + 3x^2 + 4x^3 + \dots, \quad \text{for } |x| < 1. \]

To evaluate its sum, we first examine the related geometric series:

\[ x + x^2 + x^3 + x^4 + \dots. \]

Since this is an infinite geometric progression with first term \( a = x \) and common ratio \( r = x \), its sum, valid for \( |x| < 1 \), is given by the standard formula for the sum of an infinite geometric series:

\[ \sum_{n=0}^{\infty} a r^n = \frac{a}{1 - r}, \quad \text{for } |r| < 1. \]

Applying this to our series with \( a = x \) and \( r = x \):

\[ x + x^2 + x^3 + x^4 + \dots = \frac{x}{1 - x}, \quad \text{for } |x| < 1. \]

Differentiating both sides with respect to \( x \):

\[ \frac{d}{dx} \left( x + x^2 + x^3 + x^4 + \dots \right) = \frac{d}{dx} \left( \frac{x}{1-x} \right). \]

Applying term-by-term differentiation on the left:

\[ 1 + 2x + 3x^2 + 4x^3 + \dots. \]

Using the quotient rule on the right:

\[ \frac{d}{dx} \left( \frac{x}{1-x} \right) = \frac{(1-x)(1) - x(-1)}{(1-x)^2} = \frac{1}{(1-x)^2}. \]

Thus, the sum of the given series is:

\[ 1 + 2x + 3x^2 + 4x^3 + \dots = \frac{1}{(1-x)^2}, \quad \text{for } |x| < 1. \]

Consider another infinite series as example:

\[ 1^2 + 2^2 x + 3^2 x^2 + 4^2 x^3 + \dots, \quad \text{for } |x| < 1. \]

To evaluate its sum, we begin with the geometric series:

\[ 1 + x + x^2 + x^3 + x^4 + \dots = \frac{1}{1-x}, \quad \text{for } |x| < 1. \]

Differentiating both sides with respect to \( x \),

\[ \frac{d}{dx} \left( 1 + x + x^2 + x^3 + x^4 + \dots \right) = \frac{d}{dx} \left( \frac{1}{1-x} \right). \]

Using term-by-term differentiation on the left,

\[ 0 + 1 + 2x + 3x^2 + 4x^3 + \dots. \]

Applying the derivative to the right-hand side,

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

Thus,

\[ 1 + 2x + 3x^2 + 4x^3 + \dots = \frac{1}{(1-x)^2}, \quad \text{for } |x| < 1. \]

Multiplying both sides by \( x \),

\[ x + 2x^2 + 3x^3 + 4x^4 + \dots = \frac{x}{(1-x)^2}. \]

Differentiating again,

\[ \frac{d}{dx} \left( x + 2x^2 + 3x^3 + 4x^4 + \dots \right) = \frac{d}{dx} \left( \frac{x}{(1-x)^2} \right). \]

Applying term-by-term differentiation on the left,

\[ 1 + 2 \cdot 2x + 3 \cdot 3x^2 + 4 \cdot 4x^3 + \dots. \]

Using the quotient rule on the right,

\[ \frac{(1-x)^2 (1) - x (2(1-x)(-1))}{(1-x)^4}. \]

Simplifying,

\[ \frac{(1-x)^2 + 2x(1-x)}{(1-x)^4} = \frac{(1-x+2x)(1-x)}{(1-x)^4} = \frac{(1+x)(1-x)}{(1-x)^4}. \]

Canceling \( 1-x \) in the numerator,

\[ \frac{1+x}{(1-x)^3}. \]

Thus, the sum of the given series is:

\[ 1^2 + 2^2 x + 3^2 x^2 + 4^2 x^3 + \dots = \frac{1+x}{(1-x)^3}, \quad \text{for } |x| < 1. \]

Consider the following series as another example:

\[ \cos x + 2\cos 2x + 3\cos 3x + \dots + n\cos nx. \]

To find its closed-form sum, we start with the known result for the sum of sine terms in an arithmetic sequence:

\[ \sum_{k=1}^{n} \sin kx = \frac{\sin \frac{nx}{2}}{\sin \frac{x}{2}} \sin \frac{(n+1)x}{2}. \]

This follows from the standard summation formula:

\[ \sum_{k=0}^{n-1} \sin (a + kd) = \frac{\sin \frac{nd}{2}}{\sin \frac{d}{2}} \sin \left( a + \frac{(n-1)d}{2} \right). \]

Setting \( a = x \) and \( d = x \), we obtain:

\[ \sum_{k=1}^{n} \sin kx = \frac{\sin \frac{nx}{2}}{\sin \frac{x}{2}} \sin \frac{(n+1)x}{2}. \]

Differentiating both sides with respect to \( x \),

\[ \sum_{k=1}^{n} k \cos kx = \frac{d}{dx} \left[ \frac{\sin \frac{nx}{2}}{\sin \frac{x}{2}} \sin \frac{(n+1)x}{2} \right]. \]

This derivative can be computed explicitly. I leave it as exercise. The final result provides a closed-form expression for the given summation.