Maxima Minima & Leibnitz Notes FE Sem 1 Maths-I | Mumbai Univ

This module has two parts. First, partial derivatives are used to locate and classify stationary points of a function of two independent variables, the basis of optimisation problems in design and control. Second, successive differentiation gives formulas for the nth derivative of standard functions, and Leibnitz's theorem gives the nth derivative of a product, which is used to build recurrence relations among derivatives.

Summary

Maxima and minima of f(x, y) (4.1): f has a maximum at (a, b) if f(a, b) is greater than f at all nearby points, and a minimum if it is smaller. At such an interior point, if the partial derivatives exist, \( f_x = 0 \) and \( f_y = 0 \); points satisfying both equations are called stationary (critical) points. These conditions are necessary but not sufficient.

Classification (Lagrange's conditions): at a stationary point let \( r = f_{xx}, s = f_{xy}, t = f_{yy} \). If \( rt - s^2 > 0 \) and r < 0, f is maximum; if \( rt - s^2 > 0 \) and r > 0, f is minimum; if \( rt - s^2 < 0 \), the point is a saddle point (neither maximum nor minimum); if \( rt - s^2 = 0 \), the test fails and further investigation is needed.

Working procedure: find \( f_x, f_y \); solve \( f_x = 0, f_y = 0 \) simultaneously for all stationary points; compute r, s, t; test each point; finally evaluate f at the maximum and minimum points to get the extreme values. Factorising the equations carefully is essential so that no stationary point is missed.

Successive differentiation (4.2): \( y_n = \frac{d^ny}{dx^n} \) is the nth derivative. Standard results: \( D^n(ax + b)^m = m(m - 1)\cdots(m - n + 1)a^n(ax + b)^{m-n} \); \( D^n\frac{1}{ax + b} = \frac{(-1)^nn!\,a^n}{(ax + b)^{n+1}} \); \( D^n\log(ax + b) = \frac{(-1)^{n-1}(n - 1)!\,a^n}{(ax + b)^n} \); \( D^ne^{ax} = a^ne^{ax} \); \( D^n\sin(ax + b) = a^n\sin\left(ax + b + \frac{n\pi}{2}\right) \); \( D^n\cos(ax + b) = a^n\cos\left(ax + b + \frac{n\pi}{2}\right) \).

Exponential times sine or cosine: \( D^n[e^{ax}\sin(bx + c)] = r^ne^{ax}\sin(bx + c + n\phi) \) and \( D^n[e^{ax}\cos(bx + c)] = r^ne^{ax}\cos(bx + c + n\phi) \), where \( r = \sqrt{a^2 + b^2} \) and \( \phi = \tan^{-1}\frac ba \). Rational functions are first split into partial fractions, and powers or products of sines and cosines are converted into sums using trigonometric identities, before applying the standard formulas.

Leibnitz's theorem (statement, without proof): if u and v are functions of x with nth derivatives, then \( (uv)_n = u_nv + {}^nC_1u_{n-1}v_1 + {}^nC_2u_{n-2}v_2 + \cdots + uv_n \). Choose v as the factor whose derivatives vanish quickly (a polynomial) so that the series terminates.

A standard application is to obtain a relation between \( y_{n+2}, y_{n+1}, y_n \): form a simple differential equation for y (e.g. \( (1 - x^2)y_2 - xy_1 = 0 \) for \( y = \sin^{-1}x \)), then differentiate it n times using Leibnitz's theorem.

Self-learning topics: Jacobians \( J = \frac{\partial(u, v)}{\partial(x, y)} = \begin{vmatrix} u_x & u_y \\ v_x & v_y \end{vmatrix} \) for two and three variables, and Lagrange's method of undetermined multipliers for finding extreme values subject to a constraint.

Mind map

Method: maxima and minima of f(x, y)

Important definitions

Formulas

Key points to remember

• f_x = 0 and f_y = 0 are necessary conditions only; always apply the rt − s² test.
• Solve the stationary-point equations by factorising, and consider every factor; missing a point is the most common error.
• The test is inconclusive when rt − s² = 0.
• For a minimum, r and t are both positive when rt − s² > 0; the sign of t agrees with r in that case.
• Break rational functions into partial fractions before finding the nth derivative.
• Convert sin³x, cos⁴x, sin x cos 2x etc. into sums of sines or cosines before differentiating n times.
• In Leibnitz's theorem, take the polynomial factor as v so that the series stops after a few terms.
• Leibnitz's theorem is to be used without proof in this syllabus.

Solved examples

Example 1. Find the extreme values of \( f(x, y) = x^3 + y^3 - 3axy \), a > 0.

\( f_x = 3x^2 - 3ay = 0 \Rightarrow y = \frac{x^2}{a} \); \( f_y = 3y^2 - 3ax = 0 \Rightarrow y^2 = ax \).

Substituting: \( \frac{x^4}{a^2} = ax \Rightarrow x(x^3 - a^3) = 0 \Rightarrow x = 0 \) or x = a. Stationary points (0, 0) and (a, a).

r = 6x, s = −3a, t = 6y.

At (0, 0): \( rt - s^2 = 0 - 9a^2 < 0 \): saddle point, no extreme value.

At (a, a): \( rt - s^2 = 36a^2 - 9a^2 = 27a^2 > 0 \) and r = 6a > 0: minimum.

Minimum value \( f(a, a) = a^3 + a^3 - 3a^3 = -a^3 \).

Example 2. Find the maximum and minimum values of \( x^3 + 3xy^2 - 15x^2 - 15y^2 + 72x \).

\( f_x = 3x^2 + 3y^2 - 30x + 72 = 0 \Rightarrow x^2 + y^2 - 10x + 24 = 0 \); \( f_y = 6xy - 30y = 6y(x - 5) = 0 \).

If y = 0: \( x^2 - 10x + 24 = 0 \Rightarrow x = 4, 6 \). If x = 5: \( 25 + y^2 - 50 + 24 = 0 \Rightarrow y = \pm1 \).

Stationary points: (4, 0), (6, 0), (5, 1), (5, −1). r = 6x − 30, s = 6y, t = 6x − 30.

(4, 0): r = −6, s = 0, t = −6, rt − s² = 36 > 0, r < 0: maximum. f(4, 0) = 64 − 240 + 288 = 112.

(6, 0): r = 6, s = 0, t = 6, rt − s² = 36 > 0, r > 0: minimum. f(6, 0) = 216 − 540 + 432 = 108.

(5, ±1): r = 0, t = 0, s = ±6, rt − s² = −36 < 0: saddle points.

Maximum value 112, minimum value 108.

Example 3. Find the nth derivative of \( y = \frac{x}{(x - 1)(x - 2)} \).

Partial fractions: \( \frac{x}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2} \), with \( A = \frac{1}{1 - 2} = -1 \), \( B = \frac{2}{2 - 1} = 2 \).

\( y = \frac{2}{x - 2} - \frac{1}{x - 1} \).

Using \( D^n\frac{1}{x - a} = \frac{(-1)^nn!}{(x - a)^{n+1}} \): \( y_n = (-1)^nn!\left[\frac{2}{(x - 2)^{n+1}} - \frac{1}{(x - 1)^{n+1}}\right] \).

Example 4. Find the nth derivative of \( \sin^3x \).

\( \sin3x = 3\sin x - 4\sin^3x \Rightarrow \sin^3x = \frac{3\sin x - \sin3x}{4} \).

\( y_n = \frac14\left[3\sin\left(x + \frac{n\pi}{2}\right) - 3^n\sin\left(3x + \frac{n\pi}{2}\right)\right] \).

Example 5. Find the nth derivative of \( y = e^{2x}\sin3x \).

Use \( D^n[e^{ax}\sin bx] = (a^2 + b^2)^{n/2}e^{ax}\sin\left(bx + n\tan^{-1}\frac ba\right) \) with a = 2, b = 3.

\( y_n = 13^{n/2}e^{2x}\sin\left(3x + n\tan^{-1}\frac32\right) \).

Example 6. Using Leibnitz's theorem, find the nth derivative of \( x^2e^{3x} \).

Let \( u = e^{3x} \) (so \( u_k = 3^ke^{3x} \)) and \( v = x^2 \) (v₁ = 2x, v₂ = 2, v₃ = 0).

\( y_n = u_nv + {}^nC_1u_{n-1}v_1 + {}^nC_2u_{n-2}v_2 = 3^ne^{3x}x^2 + n3^{n-1}e^{3x}(2x) + \frac{n(n - 1)}{2}3^{n-2}e^{3x}(2) \).

\( y_n = 3^{n-2}e^{3x}\left[9x^2 + 6nx + n(n - 1)\right] \).

Example 7. If \( y = \sin^{-1}x \), prove that \( (1 - x^2)y_{n+2} - (2n + 1)xy_{n+1} - n^2y_n = 0 \).

\( y_1 = \frac{1}{\sqrt{1 - x^2}} \Rightarrow (1 - x^2)y_1^2 = 1 \).

Differentiating: \( (1 - x^2)2y_1y_2 - 2xy_1^2 = 0 \Rightarrow (1 - x^2)y_2 - xy_1 = 0 \) (dividing by 2y₁).

Differentiate n times by Leibnitz: \( [(1 - x^2)y_{n+2} + n(-2x)y_{n+1} + \frac{n(n - 1)}{2}(-2)y_n] - [xy_{n+1} + ny_n] = 0 \).

\( (1 - x^2)y_{n+2} - (2n + 1)xy_{n+1} - (n^2 - n + n)y_n = 0 \), i.e. \( (1 - x^2)y_{n+2} - (2n + 1)xy_{n+1} - n^2y_n = 0 \).

Example 8. If \( y = e^{m\sin^{-1}x} \), prove that \( (1 - x^2)y_{n+2} - (2n + 1)xy_{n+1} - (n^2 + m^2)y_n = 0 \).

\( y_1 = \frac{my}{\sqrt{1 - x^2}} \Rightarrow (1 - x^2)y_1^2 = m^2y^2 \).

Differentiating: \( (1 - x^2)2y_1y_2 - 2xy_1^2 = 2m^2yy_1 \Rightarrow (1 - x^2)y_2 - xy_1 - m^2y = 0 \).

Leibnitz n times: \( (1 - x^2)y_{n+2} - 2nxy_{n+1} - n(n - 1)y_n - xy_{n+1} - ny_n - m^2y_n = 0 \).

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

Important questions and answers

Q1. State the conditions for maxima, minima and saddle point of f(x, y). (3 marks)

Stationary points satisfy \( f_x = 0, f_y = 0 \). With \( r = f_{xx}, s = f_{xy}, t = f_{yy} \) at such a point:

\( rt - s^2 > 0 \) and r < 0: maximum; \( rt - s^2 > 0 \) and r > 0: minimum; \( rt - s^2 < 0 \): saddle point; \( rt - s^2 = 0 \): no conclusion, further investigation needed.

Q2. Find the stationary points of \( f(x, y) = x^2 + y^2 + 6x + 12 \) and its extreme value. (3 marks)

\( f_x = 2x + 6 = 0 \Rightarrow x = -3 \); \( f_y = 2y = 0 \Rightarrow y = 0 \). Stationary point (−3, 0).

r = 2, s = 0, t = 2; rt − s² = 4 > 0, r > 0: minimum.

Minimum value f(−3, 0) = 9 − 18 + 12 = 3.

Q3. Examine \( f(x, y) = xy(a - x - y) \) for extreme values, a > 0. (6 marks)

\( f = axy - x^2y - xy^2 \). \( f_x = y(a - 2x - y) = 0 \), \( f_y = x(a - x - 2y) = 0 \).

Cases: y = 0, x = 0 gives (0, 0); y = 0, x + 2y = a gives (a, 0); x = 0, 2x + y = a gives (0, a); 2x + y = a and x + 2y = a give (a/3, a/3).

r = −2y, s = a − 2x − 2y, t = −2x.

(0, 0): r = 0, t = 0, s = a, rt − s² = −a² < 0: saddle. (a, 0): r = 0, t = −2a, s = −a, rt − s² = −a² < 0: saddle. (0, a): r = −2a, t = 0, s = −a, rt − s² = −a² < 0: saddle.

(a/3, a/3): r = −2a/3, t = −2a/3, s = −a/3, rt − s² = 4a²/9 − a²/9 = a²/3 > 0, r < 0: maximum.

Maximum value \( \frac a3\cdot\frac a3\left(a - \frac{2a}{3}\right) = \frac{a^3}{27} \).

Q4. Find the nth derivative of \( \frac{1}{x^2 - 5x + 6} \). (3 marks)

\( \frac{1}{(x - 2)(x - 3)} = \frac{1}{x - 3} - \frac{1}{x - 2} \).

\( y_n = (-1)^nn!\left[\frac{1}{(x - 3)^{n+1}} - \frac{1}{(x - 2)^{n+1}}\right] \).

Q5. Find the nth derivative of \( \cos x\cos2x\cos3x \). (4 marks)

\( \cos x\cos3x = \frac12(\cos4x + \cos2x) \), so \( y = \frac12(\cos4x\cos2x + \cos^22x) = \frac14(\cos6x + \cos2x) + \frac14(1 + \cos4x) \).

\( y = \frac14[1 + \cos2x + \cos4x + \cos6x] \).

\( y_n = \frac14\left[2^n\cos\left(2x + \frac{n\pi}{2}\right) + 4^n\cos\left(4x + \frac{n\pi}{2}\right) + 6^n\cos\left(6x + \frac{n\pi}{2}\right)\right] \) (n ≥ 1).

Q6. Find the nth derivative of \( \log(ax + x^2) \). (3 marks)

\( \log(ax + x^2) = \log x + \log(x + a) \).

\( y_n = (-1)^{n-1}(n - 1)!\left[\frac{1}{x^n} + \frac{1}{(x + a)^n}\right] \).

Q7. State Leibnitz's theorem and use it to find \( y_5 \) for \( y = x^3\log x \). (4 marks)

Statement: \( (uv)_n = u_nv + {}^nC_1u_{n-1}v_1 + \cdots + {}^nC_ru_{n-r}v_r + \cdots + uv_n \).

Take \( u = \log x \), \( v = x^3 \): v₁ = 3x², v₂ = 6x, v₃ = 6, v₄ = 0. \( u_k = \frac{(-1)^{k-1}(k - 1)!}{x^k} \): u₂ = −1/x², u₃ = 2/x³, u₄ = −6/x⁴, u₅ = 24/x⁵.

\( y_5 = u_5v + 5u_4v_1 + 10u_3v_2 + 10u_2v_3 = \frac{24}{x^5}x^3 + 5\left(-\frac{6}{x^4}\right)3x^2 + 10\frac{2}{x^3}6x + 10\left(-\frac{1}{x^2}\right)6 \).

\( = \frac{24 - 90 + 120 - 60}{x^2} = -\frac{6}{x^2} \).

Check: \( y_3 = 6\log x + 11 \), so \( y_4 = 6/x \), \( y_5 = -6/x^2 \).

Q8. If \( y = \cos(m\log x) \), prove that \( x^2y_{n+2} + (2n + 1)xy_{n+1} + (n^2 + m^2)y_n = 0 \). (5 marks)

\( y_1 = -\sin(m\log x)\frac mx \Rightarrow xy_1 = -m\sin(m\log x) \).

Differentiating: \( xy_2 + y_1 = -m^2\cos(m\log x)\frac1x \Rightarrow x^2y_2 + xy_1 + m^2y = 0 \).

Leibnitz n times: \( x^2y_{n+2} + n(2x)y_{n+1} + \frac{n(n - 1)}{2}(2)y_n + xy_{n+1} + ny_n + m^2y_n = 0 \).

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

Q9. Find the nth derivative of \( e^{ax}\cos(bx + c) \). (4 marks)

\( y_1 = ae^{ax}\cos(bx + c) - be^{ax}\sin(bx + c) \). Put a = r cos φ, b = r sin φ, so \( r = \sqrt{a^2 + b^2} \), \( \phi = \tan^{-1}\frac ba \).

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

Repeating, \( y_n = r^ne^{ax}\cos(bx + c + n\phi) = (a^2 + b^2)^{n/2}e^{ax}\cos\left(bx + c + n\tan^{-1}\frac ba\right) \).

Exam tips

• Show the full table of stationary points with r, s, t, rt − s² and the conclusion; examiners award marks for each point.
• Do not forget to state the maximum or minimum value, not just the point.
• In nth-derivative problems, first reduce the function (partial fractions, product-to-sum, log rules) and say so.
• For Leibnitz problems, list u_k and v_k before substituting.
• In recurrence proofs, first get a first or second order differential equation free of radicals by squaring, then apply Leibnitz.
• Write nC₂ as n(n − 1)/2 and collect coefficients carefully; the final coefficient of y_n is where most mistakes occur.

FAQs

What happens when rt − s² = 0?

The second-derivative test fails, and the nature of the point has to be decided by examining the sign of f(a + h, b + k) − f(a, b) for small h and k. MU questions normally avoid this case.

Which factor should be taken as v in Leibnitz's theorem?

Take the factor whose derivatives become zero soonest, usually a polynomial like x² or (1 − x²); then only a few terms of the series survive.

Are Jacobians and Lagrange's multipliers examined?

They are self-learning topics in the NEP syllabus, so simple problems may appear, but the main focus is the rt − s² test, nth derivatives and Leibnitz's theorem.

Why do we need partial fractions for nth derivatives?

There is no simple formula for the nth derivative of a general rational function, but there is one for 1/(ax + b). Splitting into partial fractions lets each term be handled with that formula.

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Applied Mathematics-I) and were prepared with AI help and checked.