S.M. Yusuf Calculus Chapter 7: Partial Differentiation & Euler's Theorem Solved Problems

1. Introduction to Partial Differentiation

In real-world mathematical modeling, quantities often depend on more than one independent variable. Partial differentiation extends single-variable calculus to multi-variable functions by differentiating with respect to one variable while holding all other variables constant.

Let $z = f(x, y)$ be a function of two independent variables $x$ and $y$.

First-Order Partial Derivatives

, The partial derivative of $f$ with respect to $x$ at a point $(x, y)$ is defined as the limit:

$$\frac{\partial z}{\partial x} = f_x(x, y) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x, y) - f(x, y)}{\Delta x}$$

Similarly, the partial derivative of $f$ with respect to $y$ is defined as:

$$\frac{\partial z}{\partial y} = f_y(x, y) = \lim_{\Delta y \to 0} \frac{f(x, y + \Delta y) - f(x, y)}{\Delta y}$$

Higher-Order Partial Derivatives

For a function $z = f(x, y)$, the second-order partial derivatives are given by:

• $\frac{\partial^2 z}{\partial x^2} = f_{xx} = \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)$
• $\frac{\partial^2 z}{\partial y^2} = f_{yy} = \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right)$
• $\frac{\partial^2 z}{\partial x \partial y} = f_{yx} = \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)$
• $\frac{\partial^2 z}{\partial y \partial x} = f_{xy} = \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)$

Note (Euler-Schwarz Theorem): If $f(x, y)$ and its partial derivatives $f_x, f_y, f_{xy}, f_{yx}$ are continuous on an open region, then mixed partial derivatives are equal:

$$\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y} \quad \text{or} \quad f_{xy} = f_{yx}$$ ---

2. Homogeneous Functions and Euler's Theorem

Definition of Homogeneous Function

A function $f(x, y)$ is said to be a homogeneous function of degree $n$ in $x$ and $y$ if for any real number $t > 0$:

$$f(tx, ty) = t^n f(x, y)$$

Alternatively, a homogeneous function of degree $n$ can always be written in the form:

$$f(x, y) = x^n \phi\left(\frac{y}{x}\right) \quad \text{or} \quad f(x, y) = y^n \psi\left(\frac{x}{y}\right)$$

Euler's Theorem on Homogeneous Functions

Statement: If $u = f(x, y)$ is a homogeneous function of degree $n$ possessing continuous first-order partial derivatives, then:

$$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n u$$

Extension to Three Variables:

If $u = f(x, y, z)$ is a homogeneous function of degree $n$, then:

$$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} = n u$$

Deductions / Corollaries of Euler's Theorem

Deduction 1: If $u = f(x, y)$ is not homogeneous, but $z = F(u)$ is a homogeneous function of degree $n$, then:

$$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n \frac{F(u)}{F'(u)}$$

Deduction 2 (Second-Order Form): If $u = f(x, y)$ is a homogeneous function of degree $n$, then:

$$x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = n(n - 1) u$$ ---

3. Fully Solved Step-by-Step Textbook Examples (S.M. Yusuf Ch 7)

Problem 1:

If $u = \ln(x^3 + y^3 + z^3 - 3xyz)$, show that:

$$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = \frac{3}{x + y + z}$$

Solution:

Given $u = \ln(x^3 + y^3 + z^3 - 3xyz)$. Differentiating partially with respect to $x$ while treating $y$ and $z$ as constants:

$$\frac{\partial u}{\partial x} = \frac{1}{x^3 + y^3 + z^3 - 3xyz} \cdot \frac{\partial}{\partial x}(x^3 + y^3 + z^3 - 3xyz)$$ $$\frac{\partial u}{\partial x} = \frac{3x^2 - 3yz}{x^3 + y^3 + z^3 - 3xyz}$$

By symmetry, the partial derivatives with respect to $y$ and $z$ are:

$$\frac{\partial u}{\partial y} = \frac{3y^2 - 3xz}{x^3 + y^3 + z^3 - 3xyz}$$ $$\frac{\partial u}{\partial z} = \frac{3z^2 - 3xy}{x^3 + y^3 + z^3 - 3xyz}$$

Adding the three equations together:

$$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = \frac{(3x^2 - 3yz) + (3y^2 - 3xz) + (3z^2 - 3xy)}{x^3 + y^3 + z^3 - 3xyz}$$ $$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = \frac{3(x^2 + y^2 + z^2 - xy - yz - zx)}{x^3 + y^3 + z^3 - 3xyz}$$

Using the algebraic identity $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$:

$$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = \frac{3(x^2 + y^2 + z^2 - xy - yz - zx)}{(x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)}$$ $$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = \frac{3}{x + y + z}$$

(Hence Proved)

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Problem 2:

If $u = \sin^{-1}\left(\frac{x + y}{\sqrt{x} + \sqrt{y}}\right)$, prove that:

$$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{1}{2} \tan u$$

Solution:

Rewrite the equation as:

$$z = \sin u = \frac{x + y}{\sqrt{x} + \sqrt{y}}$$

To check if $z$ is a homogeneous function, replace $x$ with $tx$ and $y$ with $ty$:

$$z(tx, ty) = \frac{tx + ty}{\sqrt{tx} + \sqrt{ty}} = \frac{t(x + y)}{\sqrt{t}(\sqrt{x} + \sqrt{y})} = t^{1 - 1/2} \frac{x + y}{\sqrt{x} + \sqrt{y}} = t^{1/2} z(x, y)$$

Thus, $z = \sin u$ is a homogeneous function of degree $n = \frac{1}{2}$ in $x$ and $y$.

By Euler's Theorem on Homogeneous Functions for $z$:

$$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = n z$$ $$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = \frac{1}{2} \sin u$$

Since $z = \sin u$, using the chain rule:

$$\frac{\partial z}{\partial x} = \cos u \frac{\partial u}{\partial x} \quad \text{and} \quad \frac{\partial z}{\partial y} = \cos u \frac{\partial u}{\partial y}$$

Substitute these into Euler's equation:

$$x \left(\cos u \frac{\partial u}{\partial x}\right) + y \left(\cos u \frac{\partial u}{\partial y}\right) = \frac{1}{2} \sin u$$ $$\cos u \left(x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}\right) = \frac{1}{2} \sin u$$ $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{1}{2} \frac{\sin u}{\cos u} = \frac{1}{2} \tan u$$

(Hence Proved)

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Problem 3:

If $u = \tan^{-1}\left(\frac{x^3 + y^3}{x - y}\right)$, prove that:

$$x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = \sin 2u \left(1 - 4 \sin^2 u\right)$$

Solution:

Let $z = \tan u = \frac{x^3 + y^3}{x - y}$.

Checking homogeneity: degree of numerator is 3, degree of denominator is 1. Thus, degree $n = 3 - 1 = 2$.

Using deduction formula for first derivatives: $x u_x + y u_y = g(u)$, where $g(u) = n \frac{F(u)}{F'(u)}$.

$$g(u) = 2 \cdot \frac{\tan u}{\sec^2 u} = 2 \sin u \cos u = \sin 2u$$

Thus, $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \sin 2u$.

Using Deduction 2 for second-order derivatives when $x u_x + y u_y = g(u)$:

$$x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = g(u)[g'(u) - 1]$$

Here $g(u) = \sin 2u$, so $g'(u) = 2 \cos 2u$. Substituting these values:

$$x^2 u_{xx} + 2xy u_{xy} + y^2 u_{yy} = \sin 2u (2 \cos 2u - 1)$$

Using the identity $\cos 2u = 1 - 2 \sin^2 u$:

$$2 \cos 2u - 1 = 2(1 - 2 \sin^2 u) - 1 = 2 - 4 \sin^2 u - 1 = 1 - 4 \sin^2 u$$

Substitute back:

$$x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = \sin 2u \left(1 - 4 \sin^2 u\right)$$

(Hence Proved)

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Problem 4:

If $u = x f\left(\frac{y}{x}\right) + g\left(\frac{y}{x}\right)$, prove that:

$$x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = 0$$

Solution:

Let $u = v + w$, where $v = x f\left(\frac{y}{x}\right)$ and $w = g\left(\frac{y}{x}\right) = x^0 g\left(\frac{y}{x}\right)$.

• $v$ is a homogeneous function of degree $n_1 = 1$.
• $w$ is a homogeneous function of degree $n_2 = 0$.

Applying Second-Order Euler's Theorem to $v$ ($n = 1$):

$$x^2 v_{xx} + 2xy v_{xy} + y^2 v_{yy} = 1(1 - 1) v = 0$$

Applying Second-Order Euler's Theorem to $w$ ($n = 0$):

$$x^2 w_{xx} + 2xy w_{xy} + y^2 w_{yy} = 0(0 - 1) w = 0$$

Since $u = v + w$, by linearity of differential operators:

$$x^2 u_{xx} + 2xy u_{xy} + y^2 u_{yy} = (x^2 v_{xx} + 2xy v_{xy} + y^2 v_{yy}) + (x^2 w_{xx} + 2xy w_{xy} + y^2 w_{yy}) = 0 + 0 = 0$$

(Hence Proved)

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4. Solved Multiple Choice Questions (MCQs)

Essential practice for University Mid/Final Term exams, PPSC Mathematics Lecturer tests, and NTS Subject tests.

1. What is the degree of homogeneity of the function $f(x, y) = \frac{x^4 + y^4}{x - y}$?
   (A) 1
   (B) 2
   (C) 3
   (D) 4
   Answer: (C) 3
   Explanation: $f(tx, ty) = \frac{t^4(x^4 + y^4)}{t(x - y)} = t^{4-1} f(x, y) = t^3 f(x, y)$. Degree $n = 3$.


2. If $u = f(x, y)$ is a homogeneous function of degree $n$, then $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$ equals:
   (A) $n u$
   (B) $(n-1) u$
   (C) $n(n-1) u$
   (D) 0
   Answer: (A) $n u$
   Explanation: Direct statement of Euler's Theorem for homogeneous functions.


3. If $f(x, y) = x^y$, what is $\frac{\partial f}{\partial x}$?
   (A) $y x^{y-1}$
   (B) $x^y \ln x$
   (C) $y x^y$
   (D) $x^{y-1}$
   Answer: (A) $y x^{y-1}$
   Explanation: When differentiating partially with respect to $x$, $y$ is treated as a constant exponent. Hence $\frac{\partial}{\partial x}(x^y) = y x^{y-1}$.


4. For $u = \frac{x^{1/3} + y^{1/3}}{x^{1/4} + y^{1/4}}$, the value of $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$ is:
   (A) $\frac{1}{12} u$
   (B) $\frac{1}{7} u$
   (C) $\frac{7}{12} u$
   (D) $\frac{1}{3} u$
   Answer: (A) $\frac{1}{12} u$
   Explanation: Degree $n = \frac{1}{3} - \frac{1}{4} = \frac{4 - 3}{12} = \frac{1}{12}$. By Euler's Theorem, the value is $n u = \frac{1}{12} u$.


5. If $f(x, y)$ has continuous second-order partial derivatives, then $f_{xy} = f_{yx}$ is guaranteed by:
   (A) Rolle's Theorem
   (B) Clairaut's / Schwarz's Theorem
   (C) Taylor's Theorem
   (D) Mean Value Theorem
   Answer: (B) Clairaut's / Schwarz's Theorem
   Explanation: Schwarz/Clairaut theorem guarantees equality of mixed partial derivatives under continuous conditions.


6. If $u = \cos\left(\frac{x}{y}\right)$, what is the degree of homogeneity $n$?
   (A) 1
   (B) -1
   (C) 0
   (D) Undefined
   Answer: (C) 0
   Explanation: $u(tx, ty) = \cos\left(\frac{tx}{ty}\right) = \cos\left(\frac{x}{y}\right) = t^0 u(x, y)$. Degree $n = 0$.


7. If $u$ is homogeneous of degree $n$, then $x^2 u_{xx} + 2xy u_{xy} + y^2 u_{yy} =$
   (A) $n u$
   (B) $n(n-1) u$
   (C) $n^2 u$
   (D) $(n+1) u$
   Answer: (B) $n(n-1) u$
   Explanation: This is the standard second-order derivative deduction of Euler's Theorem.


8. If $z = e^{xy}$, then $\frac{\partial^2 z}{\partial x \partial y}$ at $(1, 0)$ is equal to:
   (A) 0
   (B) 1
   (C) $e$
   (D) 2
   Answer: (B) 1
   Explanation: $\frac{\partial z}{\partial y} = x e^{xy}$. Then $\frac{\partial}{\partial x}\left(x e^{xy}\right) = e^{xy} + x y e^{xy} = e^{xy}(1 + xy)$. At $(1, 0)$, $e^0 (1 + 0) = 1$.