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S.M. Yusuf Calculus Chapter 7: Partial Differentiation & Euler's Theorem Solved Problems
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S.M. Yusuf Calculus Chapter 7: Partial Differentiation & Euler's Theorem Solved Problems
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 $y$.
, 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:
$$\frac{\partial z}{\partial y} = f_y(x, y) = \lim_{\Delta y \to 0} \frac{f(x, y + \Delta y) - f(x, y)}{\Delta y}$$For a function $z = f(x, y)$, the second-order partial derivatives are given by:
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:
$$\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y} \quad \text{or} \quad f_{xy} = f_{yx}$$ ---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:
$$f(x, y) = x^n \phi\left(\frac{y}{x}\right) \quad \text{or} \quad f(x, y) = y^n \psi\left(\frac{x}{y}\right)$$Statement: If $u = f(x, y)$ is a homogeneous function of degree $n$ possessing continuous first-order partial derivatives, then:
$$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n u$$If $u = f(x, y, z)$ is a homogeneous function of degree $n$, then:
$$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} = n u$$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:
$$x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = n(n - 1) u$$ ---If $u = \ln(x^3 + y^3 + z^3 - 3xyz)$, show that:
$$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = \frac{3}{x + y + z}$$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)
---If $u = \sin^{-1}\left(\frac{x + y}{\sqrt{x} + \sqrt{y}}\right)$, prove that:
$$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{1}{2} \tan u$$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)
---If $u = \tan^{-1}\left(\frac{x^3 + y^3}{x - y}\right)$, prove that:
$$x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = \sin 2u \left(1 - 4 \sin^2 u\right)$$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)
---If $u = x f\left(\frac{y}{x}\right) + g\left(\frac{y}{x}\right)$, prove that:
$$x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = 0$$Let $u = v + w$, where $v = x f\left(\frac{y}{x}\right)$ and $w = g\left(\frac{y}{x}\right) = x^0 g\left(\frac{y}{x}\right)$.
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)
---Essential practice for University Mid/Final Term exams, PPSC Mathematics Lecturer tests, and NTS Subject tests.