Partial Differentiation Notes FE Sem 1 Maths-I | Mumbai Univ

Many engineering quantities depend on more than one variable, such as temperature varying with position and time. This module defines partial derivatives of first and higher order for functions of two and three variables, shows how to differentiate composite functions with the chain rule, and proves Euler's theorem for homogeneous functions of two variables. Its deductions give quick answers to otherwise lengthy problems and are a favourite MU exam question.

Summary

Functions of several variables (3.1): z = f(x, y) assigns a value to each pair (x, y), and u = f(x, y, z) to each triple. The partial derivative \( \frac{\partial z}{\partial x} \) is the ordinary derivative of z with respect to x treating y as a constant; similarly \( \frac{\partial z}{\partial y} \) treats x as constant. Common notations are \( z_x, f_x \) or p and q for the first derivatives.

Higher-order partial derivatives (3.1): differentiating again gives \( \frac{\partial^2 z}{\partial x^2} = z_{xx} \) (often r), \( \frac{\partial^2 z}{\partial y^2} = z_{yy} \) (t) and the mixed derivatives \( \frac{\partial^2 z}{\partial x\partial y} \) and \( \frac{\partial^2 z}{\partial y\partial x} \) (s). For functions whose second partial derivatives are continuous, the mixed derivatives are equal, so the order of differentiation does not matter.

Typical results proved with partial derivatives include Laplace's equation \( u_{xx} + u_{yy} = 0 \) for \( u = \log(x^2 + y^2) \), and for u = f(r) with \( r^2 = x^2 + y^2 + z^2 \), \( u_{xx} + u_{yy} + u_{zz} = f''(r) + \frac2r f'(r) \). A useful fact is \( \frac{\partial r}{\partial x} = \frac xr \).

Differentiation of composite functions (3.1): if z = f(x, y) with x = x(t), y = y(t), then the total derivative is \( \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \). If x = x(u, v) and y = y(u, v), then \( \frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u} \) and similarly for v. This chain rule is used for change of variables such as Cartesian to polar coordinates and for symmetric functions like u = f(x − y, y − z, z − x).

Homogeneous functions: f(x, y) is homogeneous of degree n if \( f(tx, ty) = t^nf(x, y) \) for all t > 0, or equivalently it can be written as \( x^n\phi(y/x) \) (or \( y^n\psi(x/y) \)). Example: \( x^3 + 3x^2y + y^3 \) has degree 3; \( \frac{x + y}{\sqrt x + \sqrt y} \) has degree 1/2. The degree can be negative or fractional.

Euler's theorem (3.2, with proof): if u is a homogeneous function of degree n in x and y, then \( x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = nu \). Proof: write \( u = x^n f(t) \), t = y/x; then \( u_x = nx^{n-1}f - yx^{n-2}f' \) and \( u_y = x^{n-1}f' \), so \( xu_x + yu_y = nx^nf = nu \).

Deductions from Euler's theorem (3.2, without proof): (i) \( x^2u_{xx} + 2xyu_{xy} + y^2u_{yy} = n(n - 1)u \). (ii) If u itself is not homogeneous but z = f(u) is homogeneous of degree n, then \( xu_x + yu_y = g(u) \) where \( g(u) = n\frac{f(u)}{f'(u)} \), and (iii) \( x^2u_{xx} + 2xyu_{xy} + y^2u_{yy} = g(u)[g'(u) - 1] \). These handle functions like \( u = \sin^{-1}\frac{x + y}{\sqrt x + \sqrt y} \) or \( u = \log\frac{x^4 + y^4}{x + y} \).

Self-learning topics: total differential \( du = u_xdx + u_ydy \), differentiation of implicit functions \( \frac{dy}{dx} = -\frac{f_x}{f_y} \) for f(x, y) = 0, and Euler's theorem for three variables, \( xu_x + yu_y + zu_z = nu \).

Mind map

Method: applying the Euler deduction when f(u) is homogeneous

Important definitions

Formulas

Key points to remember

• While finding ∂/∂x, every other variable is a constant; do not forget to treat y as constant inside chain-rule steps.
• For x^y: ∂/∂x gives y x^(y−1), but ∂/∂y gives x^y log x.
• Mixed partial derivatives are equal when they are continuous; use this to check your work.
• Check homogeneity by substituting tx, ty; log, sin⁻¹, tan⁻¹ of a homogeneous expression of non-zero degree is not homogeneous, so use the f(u) deduction (a function of y/x alone, such as sin⁻¹(y/x), is homogeneous of degree 0).
• Euler's theorem in the form x u_x + y u_y = n u needs u itself to be homogeneous.
• In problems with symmetric arguments like f(x − y, y − z, z − x), introduce new variables r, s, t and apply the chain rule.
• Euler's theorem for two variables is to be proved; deductions are to be used without proof.

Solved examples

Example 1. If \( u = \log(x^2 + y^2) \), prove that \( \frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = 0 \).

\( u_x = \frac{2x}{x^2 + y^2} \).

\( u_{xx} = \frac{2(x^2 + y^2) - 2x\cdot2x}{(x^2 + y^2)^2} = \frac{2(y^2 - x^2)}{(x^2 + y^2)^2} \).

By symmetry (interchange x and y), \( u_{yy} = \frac{2(x^2 - y^2)}{(x^2 + y^2)^2} \).

Adding, \( u_{xx} + u_{yy} = 0 \).

Example 2. If \( u = x^y \), verify that \( \frac{\partial^2u}{\partial x\partial y} = \frac{\partial^2u}{\partial y\partial x} \).

\( u_x = yx^{y-1} \); differentiating with respect to y (product rule): \( u_{xy} = x^{y-1} + yx^{y-1}\log x \).

\( u_y = x^y\log x \); differentiating with respect to x: \( u_{yx} = yx^{y-1}\log x + x^y\cdot\frac1x = yx^{y-1}\log x + x^{y-1} \).

Both equal \( x^{y-1}(1 + y\log x) \), so the mixed derivatives are equal.

Example 3. If u = f(r) where \( r^2 = x^2 + y^2 + z^2 \), prove that \( u_{xx} + u_{yy} + u_{zz} = f''(r) + \frac2r f'(r) \).

\( 2r\frac{\partial r}{\partial x} = 2x \Rightarrow \frac{\partial r}{\partial x} = \frac xr \).

\( u_x = f'(r)\frac xr \).

\( u_{xx} = f''(r)\frac{x^2}{r^2} + f'(r)\frac{r - x\cdot\frac xr}{r^2} = f''(r)\frac{x^2}{r^2} + f'(r)\left(\frac1r - \frac{x^2}{r^3}\right) \).

Similarly for y and z. Adding and using \( x^2 + y^2 + z^2 = r^2 \): \( \sum u_{xx} = f''(r)\frac{r^2}{r^2} + f'(r)\left(\frac3r - \frac{r^2}{r^3}\right) = f''(r) + \frac2r f'(r) \).

Example 4. If u = f(x − y, y − z, z − x), prove that \( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = 0 \).

Let r = x − y, s = y − z, t = z − x, so u = f(r, s, t).

\( u_x = f_r r_x + f_s s_x + f_t t_x = f_r(1) + f_s(0) + f_t(-1) = f_r - f_t \).

\( u_y = -f_r + f_s \), \( u_z = -f_s + f_t \).

Adding: \( u_x + u_y + u_z = 0 \).

Example 5. If z = f(x, y) with x = r cos θ, y = r sin θ, prove that \( \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2 = \left(\frac{\partial z}{\partial r}\right)^2 + \frac{1}{r^2}\left(\frac{\partial z}{\partial\theta}\right)^2 \).

Chain rule: \( z_r = z_x\cos\theta + z_y\sin\theta \).

\( z_\theta = z_x(-r\sin\theta) + z_y(r\cos\theta) \Rightarrow \frac1r z_\theta = -z_x\sin\theta + z_y\cos\theta \).

Squaring and adding: \( z_r^2 + \frac{1}{r^2}z_\theta^2 = z_x^2(\cos^2\theta + \sin^2\theta) + z_y^2(\sin^2\theta + \cos^2\theta) + 2z_xz_y(\sin\theta\cos\theta - \sin\theta\cos\theta) \).

\( = z_x^2 + z_y^2 \), as required.

Example 6. If \( u = \sin^{-1}\frac{x + y}{\sqrt x + \sqrt y} \), prove that \( xu_x + yu_y = \frac12\tan u \) and \( x^2u_{xx} + 2xyu_{xy} + y^2u_{yy} = -\frac{\sin u\cos2u}{4\cos^3u} \).

\( \sin u = \frac{x + y}{\sqrt x + \sqrt y} \). Replacing x, y by tx, ty: \( \frac{t(x + y)}{\sqrt t(\sqrt x + \sqrt y)} = t^{1/2}\sin u \), so sin u is homogeneous of degree n = 1/2.

By the deduction, \( xu_x + yu_y = g(u) = n\frac{f(u)}{f'(u)} = \frac12\frac{\sin u}{\cos u} = \frac12\tan u \).

\( g'(u) = \frac12\sec^2u \). So \( x^2u_{xx} + 2xyu_{xy} + y^2u_{yy} = g(u)[g'(u) - 1] = \frac12\tan u\left(\frac12\sec^2u - 1\right) = \frac{\tan u(\sec^2u - 2)}{4} \).

\( = \frac{\sin u}{4\cos u}\cdot\frac{1 - 2\cos^2u}{\cos^2u} = -\frac{\sin u\cos2u}{4\cos^3u} \).

Example 7. If \( u = \log\frac{x^4 + y^4}{x + y} \), show that \( xu_x + yu_y = 3 \).

\( e^u = \frac{x^4 + y^4}{x + y} \) is homogeneous of degree 4 − 1 = 3.

Euler's theorem on \( e^u \): \( x\frac{\partial e^u}{\partial x} + y\frac{\partial e^u}{\partial y} = 3e^u \).

\( e^u(xu_x + yu_y) = 3e^u \Rightarrow xu_x + yu_y = 3 \).

Example 8. If \( u = x^2 + y^2 \) with \( x = at^2 \), y = 2at, find du/dt by the chain rule and verify directly.

\( \frac{du}{dt} = u_x\frac{dx}{dt} + u_y\frac{dy}{dt} = 2x(2at) + 2y(2a) = 2at^2\cdot2at + 4at\cdot2a = 4a^2t^3 + 8a^2t \).

Direct: \( u = a^2t^4 + 4a^2t^2 \Rightarrow \frac{du}{dt} = 4a^2t^3 + 8a^2t \). Both agree.

Important questions and answers

Q1. State and prove Euler's theorem for a homogeneous function of two variables. (5 marks)

Statement: if u(x, y) is homogeneous of degree n, then \( x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = nu \).

Proof: since u is homogeneous of degree n, \( u = x^nf(t) \) where \( t = \frac yx \). Then \( \frac{\partial t}{\partial x} = -\frac{y}{x^2} \), \( \frac{\partial t}{\partial y} = \frac1x \).

\( u_x = nx^{n-1}f(t) + x^nf'(t)\left(-\frac{y}{x^2}\right) = nx^{n-1}f(t) - yx^{n-2}f'(t) \).

\( u_y = x^nf'(t)\cdot\frac1x = x^{n-1}f'(t) \).

\( xu_x + yu_y = nx^nf(t) - yx^{n-1}f'(t) + yx^{n-1}f'(t) = nx^nf(t) = nu \). Hence proved.

Q2. Verify Euler's theorem for \( u = x^3 + y^3 + 3x^2y \). (3 marks)

u is homogeneous of degree 3.

\( u_x = 3x^2 + 6xy \), \( u_y = 3y^2 + 3x^2 \).

\( xu_x + yu_y = 3x^3 + 6x^2y + 3y^3 + 3x^2y = 3(x^3 + y^3 + 3x^2y) = 3u \). Verified.

Q3. If \( u = \tan^{-1}\frac{x^3 + y^3}{x - y} \), prove that \( xu_x + yu_y = \sin2u \). (4 marks)

\( \tan u = \frac{x^3 + y^3}{x - y} \) is homogeneous of degree 3 − 1 = 2.

\( xu_x + yu_y = n\frac{f(u)}{f'(u)} = 2\frac{\tan u}{\sec^2u} = 2\sin u\cos u = \sin2u \).

Q4. For the function in the previous question, show that \( x^2u_{xx} + 2xyu_{xy} + y^2u_{yy} = 2\cos3u\sin u \). (5 marks)

g(u) = sin 2u, so g'(u) = 2 cos 2u.

\( x^2u_{xx} + 2xyu_{xy} + y^2u_{yy} = g(u)[g'(u) - 1] = \sin2u(2\cos2u - 1) \).

\( = 2\sin2u\cos2u - \sin2u = \sin4u - \sin2u = 2\cos3u\sin u \) (using sin C − sin D = 2 cos((C + D)/2) sin((C − D)/2)).

Q5. If \( u = e^{xyz} \), find \( \frac{\partial^3u}{\partial x\partial y\partial z} \). (4 marks)

\( u_z = xye^{xyz} \).

\( u_{zy} = xe^{xyz} + xy\cdot xze^{xyz} = (x + x^2yz)e^{xyz} \).

\( u_{zyx} = (1 + 2xyz)e^{xyz} + (x + x^2yz)yze^{xyz} = (1 + 3xyz + x^2y^2z^2)e^{xyz} \).

Q6. If \( u = \frac{x^2 + y^2}{x + y} \), show that \( (u_x - u_y)^2 = 4\left(1 - u_x - u_y\right) \). (5 marks)

\( u_x = \frac{2x(x + y) - (x^2 + y^2)}{(x + y)^2} = \frac{x^2 + 2xy - y^2}{(x + y)^2} \), \( u_y = \frac{y^2 + 2xy - x^2}{(x + y)^2} \).

\( u_x - u_y = \frac{2(x^2 - y^2)}{(x + y)^2} = \frac{2(x - y)}{x + y} \), so \( (u_x - u_y)^2 = \frac{4(x - y)^2}{(x + y)^2} \).

\( u_x + u_y = \frac{4xy}{(x + y)^2} \), so \( 1 - u_x - u_y = \frac{(x + y)^2 - 4xy}{(x + y)^2} = \frac{(x - y)^2}{(x + y)^2} \).

Hence \( (u_x - u_y)^2 = 4(1 - u_x - u_y) \).

Q7. If \( z = f(x, y) \), x = e^u + e^{-v}, y = e^{-u} − e^v, prove that \( \frac{\partial z}{\partial u} - \frac{\partial z}{\partial v} = x\frac{\partial z}{\partial x} - y\frac{\partial z}{\partial y} \). (5 marks)

\( x_u = e^u, x_v = -e^{-v}, y_u = -e^{-u}, y_v = -e^v \).

\( z_u = z_xe^u - z_ye^{-u} \); \( z_v = -z_xe^{-v} - z_ye^v \).

\( z_u - z_v = z_x(e^u + e^{-v}) - z_y(e^{-u} - e^v) = xz_x - yz_y \).

Q8. Define a homogeneous function and find the degree of \( \frac{\sqrt x + \sqrt y}{x + y} \) and \( x^2\tan^{-1}\frac yx \). (3 marks)

f(x, y) is homogeneous of degree n if \( f(tx, ty) = t^nf(x, y) \).

\( \frac{\sqrt{tx} + \sqrt{ty}}{tx + ty} = t^{-1/2}\frac{\sqrt x + \sqrt y}{x + y} \): degree −1/2.

\( (tx)^2\tan^{-1}\frac{ty}{tx} = t^2x^2\tan^{-1}\frac yx \): degree 2.

Q9. If \( u = x^3y^2\sin^{-1}\frac yx \), find the value of \( x^2u_{xx} + 2xyu_{xy} + y^2u_{yy} \). (3 marks)

u is homogeneous of degree 3 + 2 = 5 (sin⁻¹(y/x) has degree 0).

By the deduction, \( x^2u_{xx} + 2xyu_{xy} + y^2u_{yy} = n(n - 1)u = 20u \).

Exam tips

• Euler's theorem proof is frequently asked for 4 to 5 marks; write the statement, the form u = x^n f(y/x) and both partial derivatives neatly.
• Before applying Euler's theorem always show the homogeneity check with tx, ty and state the degree.
• Write g(u) = n f(u)/f'(u) explicitly; the second-order deduction then follows mechanically.
• Use symmetry ('similarly') to save time in u_xx + u_yy type proofs, but show at least one derivative fully.
• In chain-rule problems, list the partial derivatives of the intermediate variables first; it avoids sign slips.
• Simplify the final trigonometric expression to exactly the form asked, as marks are given for the final form.

FAQs

Is Euler's theorem for three variables in the exam?

In the NEP syllabus it is a self-learning topic. The proof is required only for two variables, but you should know the statement x u_x + y u_y + z u_z = n u.

What if the function is not homogeneous?

Look for a function of u, such as sin u, tan u or e^u, that is homogeneous and use the deduction x u_x + y u_y = n f(u)/f'(u).

Why are mixed partial derivatives equal?

When the second partial derivatives are continuous in a region, Schwarz's (Clairaut's) theorem guarantees f_xy = f_yx there. All standard exam functions satisfy this away from points where they are undefined.

How are partial derivatives used in engineering?

They describe rates of change in fields such as heat conduction, fluid flow and electromagnetism, and they appear in Laplace's, heat and wave equations.

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