S.M. Yusuf Calculus Chapter 7 Partial Differentiation & Euler Theorem Solved Problems PDF Notes - Punjab University, QAU, FBISE BS Math

Theoretical Foundations & Core Theorems

Partial differentiation extends single-variable differential calculus to functions of multiple variables. In multivariable calculus, when evaluating the rate of change of a function $z = f(x, y)$ with respect to one independent variable, all other independent variables are treated as constants.

1. Definition of Partial Derivatives

For a function of two variables $z = f(x, y)$, the first-order 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 $y$ holding $x$ constant is defined as:

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

2. Concept of Homogeneous Functions

A function $f(x, y)$ is said to be a homogeneous function of degree $n$ in variables $x$ and $y$ if for any positive real parameter $t > 0$, the following algebraic identity holds true:

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

Equivalently, any homogeneous function of degree $n$ can be expressed in the factored canonical form:

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

3. Euler's Theorem on Homogeneous Functions

Theorem Statement: If $u = f(x, y)$ is a homogeneous function of degree $n$ in $x$ and $y$, possessing continuous partial derivatives of first and second orders, 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 $n$ in $x, y, z$, then:

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

4. Generalizations & Deductions of Euler's Theorem

If $u = g(x, y)$ is not strictly homogeneous, but $f(u) = z(x, y)$ is a homogeneous function of degree $n$ in $x$ and $y$, the following extended forms apply:

First-Order Extension:

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

Second-Order Extension:

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

5. Total Differential

For a function $u = f(x, y, z)$, the total differential $du$ represents the principal linear part of the total change in $u$ resulting from independent changes $dx, dy, dz$ in the variables:

$$du = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy + \frac{\partial u}{\partial z} dz$$

Formula Summary & Quick Reference

Key Concept	Formula / Mathematical Identity	Application / Significance
Homogeneity Test	$f(tx, ty) = t^n f(x, y)$	Determines the degree of homogeneity $n$ of a function.
First-Order Euler's Theorem	$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n u$	Evaluates first-order linear partial differential expressions for homogeneous $u$.
Extended First-Order Euler Form	$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n \frac{f(u)}{f'(u)}$	Used when $f(u)$ is homogeneous of degree $n$, but $u$ itself is non-homogeneous.
Second-Order Euler Relation	$x^2 u_{xx} + 2xy u_{xy} + y^2 u_{yy} = n(n-1)u$	Directly evaluates second-order homogeneous quadratic differential combinations.
Extended Second-Order Form	$x^2 u_{xx} + 2xy u_{xy} + y^2 u_{yy} = F(u)[F'(u) - 1]$	Computes second-order partial equations when $f(u)$ is homogeneous.
Total Differential Formula	$du = f_x dx + f_y dy + f_z dz$	Used in error approximations, small variations, and exact differentials.

Step-by-Step Solved Board Exam Questions

Problem 1 (Punjab University / FBISE Past Paper)

If $u = \sin^{-1} \left( \frac{x^2 + y^2}{x + y} \right)$, prove that:

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

Detailed Mathematical Proof:

Step 1: Transform to a Homogeneous Form.
Given $u = \sin^{-1} \left( \frac{x^2 + y^2}{x + y} \right)$. Taking the sine of both sides:

$$\sin u = \frac{x^2 + y^2}{x + y}$$

Let $z = \sin u = f(x, y) = \frac{x^2 + y^2}{x + y}$.

Step 2: Determine the Degree of Homogeneity.
Replace $x$ with $tx$ and $y$ with $ty$:

$$f(tx, ty) = \frac{(tx)^2 + (ty)^2}{tx + ty} = \frac{t^2 (x^2 + y^2)}{t(x + y)} = t^1 \left( \frac{x^2 + y^2}{x + y} \right) = t^1 f(x, y)$$

Therefore, $z = \sin u$ is a homogeneous function of degree $n = 1$ in $x$ and $y$.

Step 3: Apply Euler's Theorem to $z$.
By Euler's Theorem for homogeneous functions of degree $n=1$:

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

Step 4: Express Partial Derivatives of $z$ in terms of $u$.
Since $z = \sin u$:

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

Substitute these into the Euler equation:

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

Q.E.D.

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Problem 2 (UOS / KU Past Paper)

If $u = \ln \left( \frac{x^4 + y^4}{x + y} \right)$, show that:

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

Detailed Mathematical Proof:

Step 1: Transform to Exponential Form.
Given $u = \ln \left( \frac{x^4 + y^4}{x + y} \right)$. Taking the exponential on both sides:

$$e^u = \frac{x^4 + y^4}{x + y}$$

Let $z = f(u) = e^u = \frac{x^4 + y^4}{x + y}$.

Step 2: Determine Homogeneity.

$$f(tx, ty) = \frac{(tx)^4 + (ty)^4}{tx + ty} = \frac{t^4(x^4 + y^4)}{t(x + y)} = t^3 f(x, y)$$

Thus, $z = e^u$ is homogeneous of degree $n = 3$.

Step 3: First-Order Extended Euler Form.

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

Here $f(u) = e^u \implies f'(u) = e^u$. Therefore:

$$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 3 \frac{e^u}{e^u} = 3 = F(u)$$

Step 4: Apply Second-Order Extended Euler Formula.
The second-order extension identity is:

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

Since $F(u) = 3$ (a constant), $F'(u) = 0$. Substituting these values gives:

$$x^2 u_{xx} + 2xy u_{xy} + y^2 u_{yy} = 3 [0 - 1] = -3$$

Q.E.D.

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Problem 3 (QAU / PU Past Paper)

If $z = f(x, y)$ where $x = r \cos \theta$ and $y = r \sin \theta$, prove that:

$$\left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2 = \left( \frac{\partial z}{\partial r} \right)^2 + \frac{1}{r^2} \left( \frac{\partial z}{\partial \theta} \right)^2$$

Detailed Mathematical Proof:

Step 1: Apply Chain Rule for Partial Derivatives.

$$\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial r}$$ $$\frac{\partial z}{\partial \theta} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial \theta}$$

Step 2: Evaluate Partial Derivatives of Coordinate Functions.
Given $x = r \cos \theta$ and $y = r \sin \theta$:

$$\frac{\partial x}{\partial r} = \cos \theta, \quad \frac{\partial y}{\partial r} = \sin \theta$$ $$\frac{\partial x}{\partial \theta} = -r \sin \theta, \quad \frac{\partial y}{\partial \theta} = r \cos \theta$$

Step 3: Substitute and Simplify Expressions.

$$\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x} \cos \theta + \frac{\partial z}{\partial y} \sin \theta \quad \text{--- (Equation 1)}$$ $$\frac{\partial z}{\partial \theta} = -r \frac{\partial z}{\partial x} \sin \theta + r \frac{\partial z}{\partial y} \cos \theta \quad \text{--- (Equation 2)}$$

Divide Equation 2 by $r$:

$$\frac{1}{r} \frac{\partial z}{\partial \theta} = -\frac{\partial z}{\partial x} \sin \theta + \frac{\partial z}{\partial y} \cos \theta \quad \text{--- (Equation 3)}$$

Step 4: Square and Add Equations 1 and 3.

$$\left( \frac{\partial z}{\partial r} \right)^2 + \frac{1}{r^2} \left( \frac{\partial z}{\partial \theta} \right)^2 = \left( \frac{\partial z}{\partial x} \cos \theta + \frac{\partial z}{\partial y} \sin \theta \right)^2 + \left( -\frac{\partial z}{\partial x} \sin \theta + \frac{\partial z}{\partial y} \cos \theta \right)^2$$

Expanding both squares:

$$= \left( \frac{\partial z}{\partial x} \right)^2 \cos^2 \theta + \left( \frac{\partial z}{\partial y} \right)^2 \sin^2 \theta + 2 \left( \frac{\partial z}{\partial x} \right) \left( \frac{\partial z}{\partial y} \right) \cos \theta \sin \theta$$ $$+ \left( \frac{\partial z}{\partial x} \right)^2 \sin^2 \theta + \left( \frac{\partial z}{\partial y} \right)^2 \cos^2 \theta - 2 \left( \frac{\partial z}{\partial x} \right) \left( \frac{\partial z}{\partial y} \right) \sin \theta \cos \theta$$

Combining like terms using $\cos^2 \theta + \sin^2 \theta = 1$:

$$= \left( \frac{\partial z}{\partial x} \right)^2 (\cos^2 \theta + \sin^2 \theta) + \left( \frac{\partial z}{\partial y} \right)^2 (\sin^2 \theta + \cos^2 \theta)$$ $$= \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2$$

Q.E.D.

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Problem 4 (Verification of Euler's Theorem)

Verify Euler's Theorem for the function $u(x, y) = x^n \ln\left(\frac{y}{x}\right)$.

Detailed Mathematical Proof:

Step 1: Test Homogeneity.

$$u(tx, ty) = (tx)^n \ln\left(\frac{ty}{tx}\right) = t^n x^n \ln\left(\frac{y}{x}\right) = t^n u(x, y)$$

The function is homogeneous of degree $n$.

Step 2: Compute Partial Derivative $u_x$.
Using product rule on $u = x^n [\ln y - \ln x]$:

$$\frac{\partial u}{\partial x} = n x^{n-1} \ln\left(\frac{y}{x}\right) + x^n \cdot \left( -\frac{1}{x} \right) = n x^{n-1} \ln\left(\frac{y}{x}\right) - x^{n-1}$$

Multiply by $x$:

$$x \frac{\partial u}{\partial x} = n x^n \ln\left(\frac{y}{x}\right) - x^n$$

Step 3: Compute Partial Derivative $u_y$.

$$\frac{\partial u}{\partial y} = x^n \cdot \frac{1}{\left(\frac{y}{x}\right)} \cdot \left( \frac{1}{x} \right) = \frac{x^n}{y}$$

Multiply by $y$:

$$y \frac{\partial u}{\partial y} = y \left( \frac{x^n}{y} \right) = x^n$$

Step 4: Sum the Terms.

$$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \left[ n x^n \ln\left(\frac{y}{x}\right) - x^n \right] + x^n = n x^n \ln\left(\frac{y}{x}\right) = n u$$

Euler's Theorem is explicitly verified.

Interactive Practice Quiz (Clickable MCQs)

Frequently Asked Questions (FAQs)

Q1: How do I handle functions where trigonometric or inverse trigonometric functions prevent direct application of Euler's Theorem?

When functions appear in forms like $u = \sin^{-1}(f(x, y))$ or $u = \ln(f(x, y))$, $u$ itself is non-homogeneous. Transform the equation to isolate the homogeneous algebraic portion: $g(u) = f(x, y)$. Calculate the degree $n$ of $f(x, y)$, then use the extended Euler formula $x u_x + y u_y = n \frac{g(u)}{g'(u)}$.

Q2: What is the distinction between a total differential $du$ and partial derivatives $\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}$?

Partial derivatives ($\frac{\partial u}{\partial x}$) measure the rate of change of $u$ along a specific coordinate axis while keeping all other independent variables constant. In contrast, the total differential ($du = u_x dx + u_y dy$) measures the net linear approximation of the change in $u$ when all independent variables change simultaneously by small increments $dx$ and $dy$.

Q3: Why is mixed partial derivative equality ($f_{xy} = f_{yx}$) essential in university examinations?

In Pakistani university examinations (PU, UOS, QAU, KU), proving $f_{xy} = f_{yx}$ validates the continuity of second-order derivatives via Clairaut's Theorem. It serves as a test for exactness in first-order differential equations and ensures consistency when simplifying higher-order Euler relations and total differential forms.