SMTPC12 Differential Calculus : Chapter 1 - Successive Differentiation

Successive Differentiation

📑 Table of Contents

• Syllabus
• 1. Successive Differentiation
  • 1.1 What is Successive Differentiation?
  • 1.2 The n^th Derivative
• 2. Standard Results for n^th Derivatives
  • Derivative of (ax + b)^m
  • Derivative of log(ax + b)
  • Derivative of sin(ax + b)
  • Derivative of cos(ax + b)
  • Derivative of e^ax sin(bx + c)
  • Derivative of e^ax cos(bx + c)
• 3. Fractional Expressions
• 4. Trigonometric Transformation
• 5. Formation of Equations Involving Derivatives
• 6. Leibnitz Formula for the n^th Derivative of a Product
  • Formal Proof by Induction

Syllabus

Introduction (Review of basic concepts) – The n^th derivative – Standard results – Fractional expressions – Trigonometrical transformation – Formation of equations involving derivatives – Leibnitz formula for the n^th derivative of a product – Feynman’s method of differentiation.
Chapter 3 Sections 1.1 – 1.6 and Section 2.1 and 2.2

1. Successive Differentiation

1.1 What is Successive Differentiation?

When we differentiate a function, we get a new function. This new function can also be differentiated again! This process of differentiating repeatedly is called successive differentiation.

Think of differentiation like a machine:

• You put a function \(y\) in → you get its first derivative \(\dfrac{dy}{dx}\)
• Put the first derivative in → you get the second derivative \(\dfrac{d^2y}{dx^2}\)
• Put the second derivative in → you get the third derivative \(\dfrac{d^3y}{dx^3}\)
• And so on...

For a function \(y = f(x)\), the successive derivatives can be written in three different ways:

1. Prime notation: \(f'(x), \ f''(x), \ f'''(x), \ \dots, \ f^{(n)}(x)\)
2. \(y\) notation: \(y', \ y'', \ y''', \ \dots, \ y^{(n)}\)
3. Subscript notation: \(y_1, \ y_2, \ y_3, \ \dots, \ y_n\)

Example: Find successive derivatives of \(y = 4x^5\).

Solution:

Original function: \(y = 4x^5\)
First derivative: \(\dfrac{dy}{dx} = 20x^4\)
Second derivative: \(\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) = 80x^3\)
Third derivative: \(\dfrac{d}{dx}\left[\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)\right] = 240x^2\)

To make writing easier, mathematicians use shorthand:

\(\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) = \dfrac{d^2y}{dx^2} = D^2y\)
\(\dfrac{d}{dx}\left(\dfrac{d^2y}{dx^2}\right) = \dfrac{d^3y}{dx^3} = D^3y\)

In general: \(\dfrac{d}{dx}\left(\dfrac{d^{n-1}y}{dx^{n-1}}\right) = \dfrac{d^ny}{dx^n} = D^ny\)

1.2 The \(n^{th}\) Derivative

The \(n^{th}\) derivative means we differentiate the function \(n\) times. We want to find a general formula that works for any value of \(n\).

Method to Find the \(n^{th}\) Derivative:

1. Find the first few derivatives (1st, 2nd, 3rd, ...)
2. Look for a pattern
3. Guess the general formula for the \(n^{th}\) derivative
4. Verify it works for all \(n\)

Example: Find the \(n^{th}\) derivative of \(y = e^{ax}\).

Solution:

Given: \(y = e^{ax}\)
1st derivative: \(\dfrac{dy}{dx} = a e^{ax}\)
2nd derivative: \(\dfrac{d^2y}{dx^2} = a^2 e^{ax}\)
3rd derivative: \(\dfrac{d^3y}{dx^3} = a^3 e^{ax}\)

Pattern observed: Each differentiation multiplies by \(a\).

Therefore, the \(n^{th}\) derivative is: \(\dfrac{d^ny}{dx^n} = a^n e^{ax}\)

2. Standard Results for \(n^{th}\) Derivatives

Derivative of \((ax + b)^m\)
If \(y = (ax + b)^m\), then

\(y_n = m(m-1)(m-2)\cdots(m-n+1)a^n (ax + b)^{m-n}\)

Proof:

\(y_1 = m a (ax + b)^{m-1}\)
\(y_2 = m(m-1)a^2 (ax + b)^{m-2}\)
\(y_3 = m(m-1)(m-2)a^3 (ax + b)^{m-3}\)

Hence, by induction, \(y_n = m(m-1)(m-2)\cdots(m-n+1)a^n (ax + b)^{m-n}\).

Example: Find \(D^n(ax + b)^{-1}\).

Solution: Putting \(m = -1\) in the above result:

\(D^n(ax + b)^{-1} = (-1)(-2)(-3)\cdots(-n)a^n(ax + b)^{-1-n}\)
\(= (-1)^n n! a^n (ax + b)^{-n-1}\)

\(D^n(ax + b)^{-1} = (-1)^n n! a^n (ax + b)^{-n-1}\)

Derivative of \(\log(ax + b)\)
If \(y = \log(ax + b)\), then

\(y_n = (-1)^{n-1}(n-1)!a^n(ax + b)^{-n}\)

Proof:

\(y_1 = \dfrac{a}{ax + b} = a(ax + b)^{-1}\)

To find \(y_n\), we differentiate \(y_1\) \((n-1)\) times:

\(y_n = a \dfrac{d^{n-1}}{dx^{n-1}}(ax + b)^{-1}\)

Using the previous result with \(m = -1\) and replacing \(n\) by \(n-1\):

\(\dfrac{d^{n-1}}{dx^{n-1}}(ax + b)^{-1} = (-1)^{n-1}(n-1)!a^{n-1}(ax + b)^{-n}\)

Therefore,

\(y_n = a(-1)^{n-1}(n-1)!a^{n-1}(ax + b)^{-n}\)
\(= (-1)^{n-1}(n-1)!a^n(ax + b)^{-n}\)

Derivative of \(\sin(ax + b)\)
If \(y = \sin(ax + b)\), then

\(D^n\sin(ax + b) = a^n\sin\left(\dfrac{n\pi}{2} + ax + b\right)\)

Proof:

\(y_1 = a\cos(ax + b) = a\sin\left(\dfrac{\pi}{2} + ax + b\right)\)
\(y_2 = a^2\cos\left(\dfrac{\pi}{2} + ax + b\right) = a^2\sin\left(2\cdot\dfrac{\pi}{2} + ax + b\right)\)
\(y_3 = a^3\sin\left(3\cdot\dfrac{\pi}{2} + ax + b\right)\)

Each differentiation multiplies by \(a\) and increases the angle by \(\dfrac{\pi}{2}\).

Hence, by induction, \(D^n\sin(ax + b) = a^n\sin\left(\dfrac{n\pi}{2} + ax + b\right)\).

Derivative of \(\cos(ax + b)\)
If \(y = \cos(ax + b)\), then

\(D^n\cos(ax + b) = a^n\cos\left(\dfrac{n\pi}{2} + ax + b\right)\)

Proof: Starting with \(y = \cos(ax + b)\):

\(y_1 = -a\sin(ax + b) = a\cos\left(\dfrac{\pi}{2} + ax + b\right)\)

Each differentiation multiplies by \(a\) and increases the angle by \(\dfrac{\pi}{2}\).

Hence, \(D^n\cos(ax + b) = a^n\cos\left(\dfrac{n\pi}{2} + ax + b\right)\).

Corollary: Putting \(a = 1\) and \(b = 0\):

\(D^n(\sin x) = \sin\left(\dfrac{n\pi}{2} + x\right)\)
\(D^n(\cos x) = \cos\left(\dfrac{n\pi}{2} + x\right)\)

Derivative of \(e^{ax}\sin(bx + c)\)
If \(y = e^{ax}\sin(bx + c)\), then

\(D^n\{e^{ax}\sin(bx + c)\} = r^n e^{ax}\sin(bx + c + n\phi)\)

where \(r = (a^2 + b^2)^{1/2}, \quad \phi = \tan^{-1}\left(\dfrac{b}{a}\right)\)

Proof:

\(y_1 = e^{ax}b\cos(bx + c) + ae^{ax}\sin(bx + c)\)
\(= e^{ax}[b\cos(bx + c) + a\sin(bx + c)]\)

Let \(a = r\cos\phi\) and \(b = r\sin\phi\). Then \(r = \sqrt{a^2 + b^2}, \ \phi = \tan^{-1}\left(\dfrac{b}{a}\right)\).

Substituting:

\(y_1 = e^{ax}[r\sin\phi\cos(bx + c) + r\cos\phi\sin(bx + c)]\)
\(= r e^{ax}[\sin\phi\cos(bx + c) + \cos\phi\sin(bx + c)]\)
\(= r e^{ax}\sin(bx + c + \phi)\)

Now, \(y_2\) will be obtained by differentiating \(y_1\), which again multiplies by \(r\) and increases the angle by \(\phi\):

\(y_2 = r^2 e^{ax}\sin(bx + c + 2\phi)\)

Hence, by induction, \(D^n\{e^{ax}\sin(bx + c)\} = r^n e^{ax}\sin(bx + c + n\phi)\).

Derivative of \(e^{ax}\cos(bx + c)\)
If \(y = e^{ax}\cos(bx + c)\), then

\(D^n\{e^{ax}\cos(bx + c)\} = r^n e^{ax}\cos(bx + c + n\phi)\)

where \(r = (a^2 + b^2)^{1/2}, \quad \phi = \tan^{-1}\left(\dfrac{b}{a}\right)\)

Proof:

\(y_1 = e^{ax}[-b\sin(bx + c) + a\cos(bx + c)]\)
\(= e^{ax}[a\cos(bx + c) - b\sin(bx + c)]\)

With \(a = r\cos\phi\) and \(b = r\sin\phi\):

\(y_1 = e^{ax}[r\cos\phi\cos(bx + c) - r\sin\phi\sin(bx + c)]\)
\(= r e^{ax}[\cos\phi\cos(bx + c) - \sin\phi\sin(bx + c)]\)
\(= r e^{ax}\cos(bx + c + \phi)\)

By induction, \(D^n\{e^{ax}\cos(bx + c)\} = r^n e^{ax}\cos(bx + c + n\phi)\).

3. Fractional Expressions

When finding the \(n^{th}\) derivative of rational functions, we first resolve them into partial fractions. This breaks down a complicated fraction into simpler fractions whose \(n^{th}\) derivatives are known from the standard results.

Example: Find \(y_n\) where \(y = \dfrac{3}{(x+1)(2x-1)}\).

Solution: Resolving into partial fractions:

\(\dfrac{3}{(x+1)(2x-1)} = \dfrac{A}{x+1} + \dfrac{B}{2x-1}\)

To find \(A\) and \(B\): \(3 = A(2x-1) + B(x+1)\)

Putting \(x = -1\): \(3 = A(-2-1) \implies A = -1\)

Putting \(x = \dfrac{1}{2}\): \(3 = B\left(\dfrac{1}{2} + 1\right) \implies B = 2\)

Therefore, \(y = \dfrac{2}{2x-1} - \dfrac{1}{x+1}\).

Now, using the formula \(D^n(ax + b)^{-1} = (-1)^n n! a^n (ax + b)^{-n-1}\):

For \(\dfrac{1}{2x-1} = (2x-1)^{-1}\): \(D^n(2x-1)^{-1} = (-1)^n n! 2^n (2x-1)^{-n-1}\)
For \(\dfrac{1}{x+1} = (x+1)^{-1}\): \(D^n(x+1)^{-1} = (-1)^n n! (x+1)^{-n-1}\)

Therefore,

\(y_n = 2(-1)^n n! 2^n (2x-1)^{-n-1} - (-1)^n n! (x+1)^{-n-1}\)
\(= (-1)^n n! \left\{\dfrac{2^{n+1}}{(2x-1)^{n+1}} - \dfrac{1}{(x+1)^{n+1}}\right\}\)

\(y_n = (-1)^n n! \left\{\dfrac{2^{n+1}}{(2x-1)^{n+1}} - \dfrac{1}{(x+1)^{n+1}}\right\}\)

Example: Find \(y_n\) when \(y = \dfrac{x^2}{(x-1)^2(x+2)}\).

Solution: Resolving into partial fractions:

\(\dfrac{x^2}{(x-1)^2(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{C}{x+2}\)

Multiplying both sides by \((x-1)^2(x+2)\): \(x^2 = A(x-1)(x+2) + B(x+2) + C(x-1)^2\)

Put \(x=1\): \(1 = 3B \implies B = \dfrac{1}{3}\)

Put \(x=-2\): \(4 = 9C \implies C = \dfrac{4}{9}\)

Put \(x=0\): \(0 = -2A + 2B + C \implies A = \dfrac{5}{9}\)

Therefore, \(y = \dfrac{5}{9}\dfrac{1}{x-1} + \dfrac{1}{3}\dfrac{1}{(x-1)^2} + \dfrac{4}{9}\dfrac{1}{x+2}\).

Now, using standard formulas:

\(D^n(x-1)^{-1} = (-1)^n n! (x-1)^{-n-1}\)
\(D^n(x-1)^{-2} = (-1)^n (n+1)! (x-1)^{-n-2}\)
\(D^n(x+2)^{-1} = (-1)^n n! (x+2)^{-n-1}\)

Therefore,

\(y_n = (-1)^n n! \left\{\dfrac{5}{9(x-1)^{n+1}} + \dfrac{n+1}{3(x-1)^{n+2}} + \dfrac{4}{9(x+2)^{n+1}}\right\}\)

Example: Find \(y_n\) when \(y = \dfrac{1}{x^2 + a^2}\).

Solution: Note that \(x^2 + a^2 = (x + ai)(x - ai)\), where \(i = \sqrt{-1}\).
Resolving into partial fractions:

\(\dfrac{1}{x^2 + a^2} = \dfrac{1}{(x+ai)(x-ai)} = \dfrac{A}{x-ai} + \dfrac{B}{x+ai}\)

\(1 = A(x+ai) + B(x-ai)\)

Put \(x = ai\): \(A = \dfrac{1}{2ai}\)    Put \(x = -ai\): \(B = -\dfrac{1}{2ai}\)

Therefore, \(y = \dfrac{1}{2ai}\left(\dfrac{1}{x-ai} - \dfrac{1}{x+ai}\right)\).

Using \(D^n(x+c)^{-1} = (-1)^n n! (x+c)^{-n-1}\):

\(y_n = \dfrac{(-1)^n n!}{2ai}\left\{\dfrac{1}{(x-ai)^{n+1}} - \dfrac{1}{(x+ai)^{n+1}}\right\}\)

4. Trigonometric Transformation

It is possible to break up products of powers of sines and cosines into a sum by trigonometric methods. This makes it easier to find the \(n^{th}\) derivatives.

Example: Find the \(n^{th}\) differential coefficient of \(\cos x \cdot \cos 2x \cdot \cos 3x\).

Solution: Using the identity \(2\cos A\cos B = \cos(A+B) + \cos(A-B)\):

\(\cos x \cos 2x \cos 3x = \dfrac{1}{2}\cos 2x(\cos 4x + \cos 2x)\)
\(= \dfrac{1}{2}\cos 2x\cos 4x + \dfrac{1}{2}\cos^2 2x\)
\(= \dfrac{1}{4}(\cos 6x + \cos 2x) + \dfrac{1}{4}(1 + \cos 4x)\)
\(= \dfrac{1}{4} + \dfrac{1}{4}(\cos 2x + \cos 4x + \cos 6x)\)

Now, applying the \(n^{th}\) derivative formula for \(\cos(ax)\): \(D^n\cos(ax) = a^n\cos\left(\dfrac{n\pi}{2} + ax\right)\)

\(D^n(\cos x \cos 2x \cos 3x) = \dfrac{1}{4}\left\{2^n\cos\left(\dfrac{n\pi}{2} + 2x\right) + 4^n\cos\left(\dfrac{n\pi}{2} + 4x\right) + 6^n\cos\left(\dfrac{n\pi}{2} + 6x\right)\right\}\)

Example: Find the \(n^{th}\) differential coefficient of \(\cos^5\theta \sin^7\theta\).

Solution: Let \(z = \cos\theta + i\sin\theta\). Then \(\dfrac{1}{z} = \cos\theta - i\sin\theta\).
By De Moivre's Theorem:

\(z^n = \cos n\theta + i\sin n\theta\)
\(\dfrac{1}{z^n} = \cos n\theta - i\sin n\theta\)

So, \(2\cos\theta = z + \dfrac{1}{z}, \quad 2i\sin\theta = z - \dfrac{1}{z}\).

Therefore, \(2^{12}\cos^5\theta\sin^7\theta = (z + z^{-1})^5(z - z^{-1})^7(-i)^7\).

Expanding and simplifying:

\(2^{12}\cos^5\theta\sin^7\theta = -\left(\sin 12\theta - 2\sin 10\theta - 4\sin 8\theta + 10\sin 6\theta + 5\sin 4\theta - 20\sin 2\theta\right)\)

Hence,

\(\cos^5\theta\sin^7\theta = -\dfrac{1}{2^{12}}\left(\sin 12\theta - 2\sin 10\theta - 4\sin 8\theta + 10\sin 6\theta + 5\sin 4\theta - 20\sin 2\theta\right)\)

Now, applying \(D^n\sin(a\theta) = a^n\sin\left(\dfrac{n\pi}{2} + a\theta\right)\):

\(D^n(\cos^5\theta\sin^7\theta) = -\dfrac{1}{2^{11}}\left\{12^n\sin\left(\dfrac{n\pi}{2} + 12\theta\right) - 2\cdot 10^n\sin\left(\dfrac{n\pi}{2} + 10\theta\right) - 4\cdot 8^n\sin\left(\dfrac{n\pi}{2} + 8\theta\right) + 10\cdot 6^n\sin\left(\dfrac{n\pi}{2} + 6\theta\right) + 5\cdot 4^n\sin\left(\dfrac{n\pi}{2} + 4\theta\right) - 20\cdot 2^n\sin\left(\dfrac{n\pi}{2} + 2\theta\right)\right\}\)

5. Formation of Equations Involving Derivatives

When a relation between \(x\) and \(y\) is given, we can in many cases deduce from it a relation between the variables \(x\), \(y\) and the derivatives of \(y\) with respect to \(x\).

Example: If \(xy = ae^x + be^{-x}\), prove that \(x\dfrac{d^2y}{dx^2} + 2\dfrac{dy}{dx} - xy = 0\).

Solution: Given \(xy = ae^x + be^{-x}\).
Differentiating with respect to \(x\): \(y + x\dfrac{dy}{dx} = ae^x - be^{-x}\).
Differentiating again:

\(\dfrac{dy}{dx} + x\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} = ae^x + be^{-x}\)
\(x\dfrac{d^2y}{dx^2} + 2\dfrac{dy}{dx} = ae^x + be^{-x}\)

But \(ae^x + be^{-x} = xy\). Therefore, \(x\dfrac{d^2y}{dx^2} + 2\dfrac{dy}{dx} = xy\).

\(x\dfrac{d^2y}{dx^2} + 2\dfrac{dy}{dx} - xy = 0\)

Example: If \(y = \sin(m\sin^{-1}x)\), prove that \((1-x^2)y_2 - xy_1 + m^2y = 0\).

Solution: Given \(y = \sin(m\sin^{-1}x)\) ⇒ \(\sin^{-1}y = m\sin^{-1}x\).
Differentiating: \(\dfrac{1}{\sqrt{1-y^2}}\dfrac{dy}{dx} = \dfrac{m}{\sqrt{1-x^2}}\).
Squaring both sides:

\(\dfrac{1}{1-y^2}\left(\dfrac{dy}{dx}\right)^2 = \dfrac{m^2}{1-x^2}\)
\((1-x^2)\left(\dfrac{dy}{dx}\right)^2 = m^2(1-y^2)\)

Differentiating with respect to \(x\):

\((1-x^2)2\dfrac{dy}{dx}\dfrac{d^2y}{dx^2} - 2x\left(\dfrac{dy}{dx}\right)^2 = -2m^2y\dfrac{dy}{dx}\)

Cancelling the common factor \(2\dfrac{dy}{dx}\) throughout:

\((1-x^2)y_2 - xy_1 + m^2y = 0\)

Example: If \(x = \sin\theta\), \(y = \cos p\theta\), prove that \((1-x^2)y_2 - xy_1 + p^2y = 0\).

Solution: Given \(x = \sin\theta\) and \(y = \cos p\theta\).

\(\dfrac{dx}{d\theta} = \cos\theta\), \(\dfrac{dy}{d\theta} = -p\sin p\theta\)
\(\dfrac{dy}{dx} = -p\cdot\dfrac{\sin p\theta}{\cos\theta}\)

Now, \(\sin p\theta = \sqrt{1-y^2}\) and \(\cos\theta = \sqrt{1-x^2}\).

\(\dfrac{dy}{dx} = -p\dfrac{\sqrt{1-y^2}}{\sqrt{1-x^2}}\)

Squaring both sides:

\(\left(\dfrac{dy}{dx}\right)^2 = p^2\dfrac{1-y^2}{1-x^2}\)
\((1-x^2)\left(\dfrac{dy}{dx}\right)^2 = p^2(1-y^2)\)

Differentiating with respect to \(x\):

\((1-x^2)2\dfrac{dy}{dx}\dfrac{d^2y}{dx^2} - 2x\left(\dfrac{dy}{dx}\right)^2 = -2p^2y\dfrac{dy}{dx}\)

Cancelling \(2\dfrac{dy}{dx}\):

\((1-x^2)y_2 - xy_1 + p^2y = 0\)

6. Leibnitz Formula for the \(n^{th}\) Derivative of a Product

This formula expresses the \(n^{th}\) derivative of the product of two variables in terms of the variables themselves and their successive derivatives.

If \(u\) and \(v\) are functions of \(x\), we have

\(\dfrac{d}{dx}(uv) = v\dfrac{du}{dx} + u\dfrac{dv}{dx}\) i.e., \(D(uv) = vDu + uDv\).

Differentiating again with respect to \(x\):

\(D^2(uv) = D(v\cdot Du) + D(u\cdot Dv)\)
\(= vD^2u + 2Du\cdot Dv + uD^2v\)

Similarly,

\(D^3(uv) = vD^3u + 3D^2u\cdot Dv + 3Du\cdot D^2v + uD^3v\)

Continuing this process, we observe that the numerical coefficients follow the same law as that of the Binomial Theorem and the indices of the derivatives correspond to the exponents of the Binomial Theorem.

Hence,

\(\dfrac{d^n}{dx^n}(uv) = u_nv + {}^nC_1u_{n-1}v_1 + {}^nC_2u_{n-2}v_2 + \cdots + {}^nC_r u_{n-r}v_r + \cdots + uv_n\)

where \(u_n = \dfrac{d^nu}{dx^n}\) and \(v_n = \dfrac{d^nv}{dx^n}\).

Formal Proof by Induction

Proof: Assume the theorem is true for some value of \(n\), i.e., suppose

\(D^n(uv) = u_nv + {}^nC_1u_{n-1}v_1 + {}^nC_2u_{n-2}v_2 + \cdots + {}^nC_{r-1}u_{n-r+1}v_{r-1} + {}^nC_r u_{n-r}v_r + \cdots + uv_n\)

Differentiating both sides again:

\(D^{n+1}(uv) = (u_{n+1}v + u_nv_1) + {}^nC_1(u_nv_1 + u_{n-1}v_2) + {}^nC_2(u_{n-1}v_2 + u_{n-2}v_3) + \cdots + {}^nC_{r-1}(u_{n-r+2}v_{r-1} + u_{n-r+1}v_r) + {}^nC_r(u_{n-r+1}v_r + u_{n-r}v_{r+1}) + \cdots + (u_1v_n + uv_{n+1})\)

Collecting the coefficients of like terms:

\(D^{n+1}(uv) = u_{n+1}v + (1 + {}^nC_1)u_nv_1 + ({}^nC_1 + {}^nC_2)u_{n-1}v_2 + \cdots + ({}^nC_{r-1} + {}^nC_r)u_{n-r+1}v_r + \cdots + uv_{n+1}\)

Now, using the property of binomial coefficients: \({}^nC_{r-1} + {}^nC_r = {}^{n+1}C_r\)

Also, \(1 + {}^nC_1 = {}^{n+1}C_1\), \({}^nC_1 + {}^nC_2 = {}^{n+1}C_2\), etc.

Therefore,

\(D^{n+1}(uv) = u_{n+1}v + {}^{n+1}C_1u_nv_1 + {}^{n+1}C_2u_{n-1}v_2 + \cdots + {}^{n+1}C_r u_{n-r+1}v_r + \cdots + uv_{n+1}\)

Hence, if the theorem is true for any value of \(n\), it must be true for the next higher value \(n+1\). It has been seen that it is true for \(n=1\) and therefore it is true for \(n=2\), \(n=3\), and so on for all values of \(n\).

This theorem is particularly useful when one of the factors is a small integral multiple of \(x\); if this be taken as \(v\) in the preceding formula, its differential coefficients will vanish after a few terms, and the series will consist of only a few terms.

Example: Find the \(n^{th}\) differential coefficient of \(x^2\log x\).

Solution: Taking \(v = x^2\) and \(u = \log x\):

\(\dfrac{d^n}{dx^n}(x^2\log x) = \dfrac{d^n}{dx^n}(\log x)x^2 + {}^nC_1\dfrac{d^{n-1}}{dx^{n-1}}(\log x)\dfrac{d}{dx}(x^2) + {}^nC_2\dfrac{d^{n-2}}{dx^{n-2}}(\log x)\dfrac{d^2}{dx^2}(x^2)\)

All the other terms will be zero since the successive derivatives of \(x^2\) after the second derivative vanish.

Now, \(\dfrac{d^n}{dx^n}(\log x) = (-1)^{n-1}(n-1)!x^{-n}\)

\(\dfrac{d}{dx}(x^2) = 2x, \quad \dfrac{d^2}{dx^2}(x^2) = 2\)

Therefore,

\(D^n(x^2\log x) = (-1)^{n-1}(n-1)!x^{2-n} + {}^nC_1(-1)^{n-2}(n-2)!x^{1-n}\cdot 2x + {}^nC_2(-1)^{n-3}(n-3)!x^{2-n}\cdot 2\)

\(D^n(x^2\log x) = (-1)^{n-1}(n-1)!x^{2-n} + 2n(-1)^{n-2}(n-2)!x^{2-n} + n(n-1)(-1)^{n-3}(n-3)!x^{2-n}\)

Example: If \(y = \sin(m\sin^{-1}x)\), prove that \((1-x^2)y_2 - xy_1 + m^2y = 0\) and \((1-x^2)y_{n+2} - (2n+1)xy_{n+1} + (m^2 - n^2)y_n = 0\).

Solution: From the previous result \((1-x^2)y_2 = xy_1 - m^2y\).
Taking the \(n^{th}\) derivative of each term by Leibnitz's Theorem:
For the LHS: Differentiating \((1-x^2)y_2\) \(n\) times,

\(D^n[(1-x^2)y_2] = y_{n+2}(1-x^2) + {}^nC_1y_{n+1}(-2x) + {}^nC_2y_n(-2)\)
\(= (1-x^2)y_{n+2} - 2nxy_{n+1} - n(n-1)y_n\)

For the RHS: Differentiating \(xy_1 - m^2y\) \(n\) times,

\(D^n[xy_1 - m^2y] = xy_{n+1} + {}^nC_1y_n - m^2y_n = xy_{n+1} + ny_n - m^2y_n\)

Equating both sides:

\((1-x^2)y_{n+2} - 2nxy_{n+1} - n(n-1)y_n = xy_{n+1} + ny_n - m^2y_n\)

Bringing all terms to LHS:

\((1-x^2)y_{n+2} - (2n+1)xy_{n+1} + (m^2 - n^2)y_n = 0\)

\((1-x^2)y_{n+2} - (2n+1)xy_{n+1} + (m^2 - n^2)y_n = 0\)

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— End of Chapter —