S.M. Yusuf Calculus Chapter 7 Partial Differentiation & Euler's Theorem Solved Problems PDF Guide | BSc, BS Math, Punjab University, FBISE

Theoretical Foundations & Core Theorems

In multivariable calculus, partial differentiation extends the concept of single-variable derivatives to functions of several variables. When analyzing a function $f(x, y)$, the partial derivative with respect to $x$ measures the rate of change of $f$ as $x$ varies while holding $y$ strictly constant. Mathematically, the first-order partial derivatives are defined via limits as follows:

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

Geometric Interpretation

Geometrically, $z = f(x, y)$ represents a surface in three-dimensional Euclidean space $\mathbb{R}^3$. Intersecting this surface with a plane $y = y_0$ yields a space curve. The partial derivative $\frac{\partial f}{\partial x}$ evaluated at $(x_0, y_0)$ represents the slope of the tangent line to this curve at that specific point. Similarly, $\frac{\partial f}{\partial y}$ represents the slope of the tangent line along the intersection curve formed by the plane $x = x_0$.

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 real scalar $t > 0$, the following identity holds:

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

Alternatively, a homogeneous function of degree $n$ can always be written in the normalized forms $f(x, y) = x^n \phi\left(\frac{y}{x}\right)$ or $f(x, y) = y^n \psi\left(\frac{x}{y}\right)$.

Euler's Theorem on Homogeneous Functions

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

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

Proof Outline: Since $u = f(x, y)$ is homogeneous of degree $n$, we write $u = x^n \phi(v)$ where $v = \frac{y}{x}$. Differentiating $u$ partially with respect to $x$ using the product rule and chain rule:

$$\frac{\partial u}{\partial x} = n x^{n-1} \phi(v) + x^n \phi'(v) \left(-\frac{y}{x^2}\right) = n x^{n-1} \phi(v) - x^{n-1} y \phi'(v)$$

Next, differentiating $u$ partially with respect to $y$:

$$\frac{\partial u}{\partial y} = x^n \phi'(v) \left(\frac{1}{x}\right) = x^{n-1} \phi'(v)$$

Multiplying $\frac{\partial u}{\partial x}$ by $x$ and $\frac{\partial u}{\partial y}$ by $y$, and summing the expressions:

$$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = x \left[ n x^{n-1} \phi(v) - x^{n-1} y \phi'(v) \right] + y \left[ x^{n-1} \phi'(v) \right]$$ $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n x^n \phi(v) - x^n y \phi'(v) + x^{n-1} y \phi'(v) = n x^n \phi(v) = n u$$

This completes the proof. Euler's theorem extends directly to second-order partial derivatives:

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

Formula Summary & Quick Reference

Key Concept	Mathematical Formula / Condition	Academic / Exam Application
First-Order Partials	$f_x = \frac{\partial f}{\partial x}$, $f_y = \frac{\partial f}{\partial y}$	Finding local rates of change along axis planes.
Mixed Derivatives Equality (Schwarz/Clairaut)	$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$ if $f_{xy}, f_{yx}$ are continuous	Symmetry verification in higher-order differentiations.
Homogeneity Test	$f(tx, ty) = t^n f(x,y)$	Determining degree $n$ prior to applying Euler's Theorem.
First Euler Identity	$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n u$	Simplifying first-order linear partial expressions.
Second Euler Identity	$x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = n(n-1) u$	Solving second-order partial derivative combinations.
Composite Chain Rule	$\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}$	Functions dependent on parametric variables ($t, r, \theta$).

Step-by-Step Solved Board Exam Questions

Problem 1 (Punjab University / FBISE Past Paper)

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

$$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \sin(2u)$$

Solution:

Step 1: Check homogeneity of $u$.
The presence of $\tan^{-1}$ prevents $u$ from being directly homogeneous. Rewrite the equation as:

$$z = \tan(u) = \frac{x^3 + y^3}{x - y}$$

Step 2: Test degree of homogeneity for $z = f(x,y)$.

$$z(tx, ty) = \frac{(tx)^3 + (ty)^3}{tx - ty} = \frac{t^3(x^3 + y^3)}{t(x - y)} = t^{3-1} \frac{x^3 + y^3}{x - y} = t^2 z(x,y)$$

Thus, $z = \tan(u)$ is a homogeneous function of degree $n = 2$.

Step 3: Apply Euler's Theorem to $z$.
By Euler’s Theorem on homogeneous functions:

$$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = n z = 2 \tan(u)$$

Step 4: Express derivatives of $z$ in terms of $u$.
Since $z = \tan(u)$, using the chain rule gives:

$$\frac{\partial z}{\partial x} = \sec^2(u) \frac{\partial u}{\partial x}$$ $$\frac{\partial z}{\partial y} = \sec^2(u) \frac{\partial u}{\partial y}$$

Step 5: Substitute partial derivatives into Euler's relation.

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

Hence proved.

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Problem 2 (UOS / KU Past Paper)

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

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

Solution:

Step 1: Transform equation into homogeneous form.
Exponentiating both sides gives:

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

Step 2: Determine degree $n$.

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

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

Step 3: Prove Part (a).
Applying Euler's Theorem to $z$:

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

Since $z = e^u$, we have $\frac{\partial z}{\partial x} = e^u \frac{\partial u}{\partial x}$ and $\frac{\partial z}{\partial y} = e^u \frac{\partial u}{\partial y}$. Substituting these:

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

Dividing by $e^u \neq 0$:

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

Step 4: Prove Part (b).
Differentiating equation $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 3$ partially with respect to $x$:

$$\frac{\partial}{\partial x} \left( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} \right) = 0$$ $$\left( 1 \cdot \frac{\partial u}{\partial x} + x \frac{\partial^2 u}{\partial x^2} \right) + y \frac{\partial^2 u}{\partial x \partial y} = 0$$ $$x \frac{\partial^2 u}{\partial x^2} + y \frac{\partial^2 u}{\partial x \partial y} = -\frac{\partial u}{\partial x} \quad \text{--- (Equation 1)}$$

Differentiating $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 3$ partially with respect to $y$:

$$\frac{\partial}{\partial y} \left( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} \right) = 0$$ $$x \frac{\partial^2 u}{\partial y \partial x} + \left( 1 \cdot \frac{\partial u}{\partial y} + y \frac{\partial^2 u}{\partial y^2} \right) = 0$$ $$x \frac{\partial^2 u}{\partial x \partial y} + y \frac{\partial^2 u}{\partial y^2} = -\frac{\partial u}{\partial y} \quad \text{--- (Equation 2)}$$

Multiply Equation 1 by $x$ and Equation 2 by $y$, then add them:

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

Using the result from Part (a), $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 3$:

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

Hence proved.

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Problem 3 (QAU Past Paper)

Question: Verify Euler's Theorem for $u(x, y) = \frac{x^{1/4} + y^{1/4}}{x^{1/5} + y^{1/5}}$.

Solution:

Step 1: Determine homogeneity and degree $n$.

$$u(tx, ty) = \frac{(tx)^{1/4} + (ty)^{1/4}}{(tx)^{1/5} + (ty)^{1/5}} = \frac{t^{1/4}(x^{1/4} + y^{1/4})}{t^{1/5}(x^{1/5} + y^{1/5})} = t^{1/4 - 1/5} u(x, y) = t^{1/20} u(x, y)$$

Hence, $u(x, y)$ is homogeneous of degree $n = \frac{1}{20}$.

Step 2: Direct Partial Derivatives.
Let $N = x^{1/4} + y^{1/4}$ and $D = x^{1/5} + y^{1/5}$, so $u = \frac{N}{D}$.

$$\frac{\partial u}{\partial x} = \frac{D \frac{\partial N}{\partial x} - N \frac{\partial D}{\partial x}}{D^2} = \frac{D \left(\frac{1}{4}x^{-3/4}\right) - N \left(\frac{1}{5}x^{-4/5}\right)}{D^2}$$ $$\frac{\partial u}{\partial y} = \frac{D \frac{\partial N}{\partial y} - N \frac{\partial D}{\partial y}}{D^2} = \frac{D \left(\frac{1}{4}y^{-3/4}\right) - N \left(\frac{1}{5}y^{-4/5}\right)}{D^2}$$

Step 3: Construct $x u_x + y u_y$.

$$x \frac{\partial u}{\partial x} = \frac{D \left(\frac{1}{4}x^{1/4}\right) - N \left(\frac{1}{5}x^{1/5}\right)}{D^2}$$ $$y \frac{\partial u}{\partial y} = \frac{D \left(\frac{1}{4}y^{1/4}\right) - N \left(\frac{1}{5}y^{1/5}\right)}{D^2}$$

Adding the two terms:

$$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{\frac{1}{4} D \left(x^{1/4} + y^{1/4}\right) - \frac{1}{5} N \left(x^{1/5} + y^{1/5}\right)}{D^2}$$

Since $x^{1/4} + y^{1/4} = N$ and $x^{1/5} + y^{1/5} = D$:

$$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{\frac{1}{4} D N - \frac{1}{5} N D}{D^2} = \frac{\left(\frac{1}{4} - \frac{1}{5}\right) N D}{D^2} = \frac{1}{20} \frac{N}{D} = \frac{1}{20} u$$

This matches $n u$ where $n = \frac{1}{20}$. Euler's theorem is verified.

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Problem 4 (S.M. Yusuf Ch 7 Classical Derivative Question)

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

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

Solution:

Step 1: Apply Chain Rule for polar coordinates.

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

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 2: Substitute partial derivatives into chain rule equations.

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

Divide Eq 2 by $r$:

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

Step 3: Square and sum Eq 1 and Eq 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 the right side:

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

Combining like terms:

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

Since $\sin^2\theta + \cos^2\theta = 1$:

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

Hence proved.

Interactive Practice Quiz (Clickable MCQs)

Frequently Asked Questions (FAQs)

Q1: How do I handle trigonometric, inverse trigonometric, logarithmic, or exponential expressions when applying Euler's Theorem?
Functions like $\sin^{-1}\left(\frac{x^2+y^2}{x+y}\right)$ or $\ln\left(\frac{x^3+y^3}{x-y}\right)$ are not directly homogeneous because of the outer function. To solve these in university examinations, move the outer operator to the left-hand side (e.g., $z = \sin(u)$ or $z = e^u$). Then determine the degree of homogeneity $n$ of $z$. Apply Euler's Theorem to $z$, and use the chain rule ($\frac{\partial z}{\partial x} = z'(u) \frac{\partial u}{\partial x}$) to express the final result in terms of $u$.

Q2: Why is the condition of continuity essential for proving $\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$?
The equality of mixed partial derivatives (Schwarz/Clairaut Theorem) depends on the local linearity and smoothness of the surface $z = f(x, y)$. If second-order partial derivatives are discontinuous at a point $(x_0, y_0)$, the order of differentiation can affect the result, leading to unequal mixed partial derivatives at that point.

Q3: How do university examiners mark Euler's Theorem proofs?
In examinations conducted by Punjab University, UOS, KU, QAU, or FBISE, examiners award step-wise marks: 1. Correctly transforming non-homogeneous expressions into homogeneous forms (1-2 marks). 2. Explicitly identifying and demonstrating degree calculation $n$ (1-2 marks). 3. Correct application of the Chain Rule or partial derivative expansions (2-3 marks). 4. Final algebraic simplification yielding the required RHS identity (2 marks).