Answers: Introduction to partial differentiation
These are the answers to Questions: Introduction to partial differentiation.
Please attempt the questions before reading these answers!
Q1
1.1. \(\displaystyle \quad \dfrac{\partial f}{\partial x} = 2xy\) and \(\displaystyle \dfrac{\partial f}{\partial y} = x^2 + 3y^2\).
1.2. \(\displaystyle \quad \dfrac{\partial f}{\partial x} = 9x^2 + y\) and \(\displaystyle \dfrac{\partial f}{\partial y} = x-8y^3\).
1.3. \(\displaystyle \quad \dfrac{\partial f}{\partial x} = 2y \cos(2x)\) and \(\displaystyle \dfrac{\partial f}{\partial y} = \sin(2x)\).
1.4. \(\displaystyle \quad \dfrac{\partial f}{\partial x} = y e^{xy} + 4xy^3\) and \(\displaystyle \dfrac{\partial f}{\partial y} = x e^{xy} + 6x^2y^2\).
1.5. \(\displaystyle \quad \dfrac{\partial f}{\partial x} = \dfrac{1}{x} + \ln(y) + 3\) and \(\displaystyle \dfrac{\partial f}{\partial y} = \dfrac{x}{y}\).
1.6. \(\displaystyle \quad \dfrac{\partial f}{\partial x} = -\dfrac{y}{x^2} - \dfrac{1}{y}\) and \(\displaystyle \dfrac{\partial f}{\partial y} = \dfrac{1}{x} + \dfrac{x}{y^2}\).
1.7. \(\displaystyle \quad \dfrac{\partial f}{\partial x} = \exp(y^2)\) and \(\displaystyle \dfrac{\partial f}{\partial y} = 2xy \exp(y^2)\).
1.8. \(\displaystyle \quad \dfrac{\partial f}{\partial x} = \dfrac{x}{\sqrt{x^2+y^2}}\) and \(\displaystyle \dfrac{\partial f}{\partial y} = \dfrac{y}{\sqrt{x^2+y^2}}\).
1.9. \(\displaystyle \quad \dfrac{\partial f}{\partial x} = 12(3x+2y)^3\) and \(\displaystyle \dfrac{\partial f}{\partial y} = 8(3x+2y)^3\).
1.10. \(\displaystyle \quad \dfrac{\partial f}{\partial x} = y^2 \cos(xy)\) and \(\displaystyle \dfrac{\partial f}{\partial y} = x \cos(xy) - x^2y \sin(xy)\).
1.11. \(\displaystyle \quad \dfrac{\partial f}{\partial x} = 2x \cos(x^2+y^2)\) and \(\displaystyle \dfrac{\partial f}{\partial y} = 2y \cos(x^2+y^2)\).
1.12. \(\displaystyle \quad \dfrac{\partial f}{\partial x} = \dfrac{2xy^2}{1+x^2y^2}\) and \(\displaystyle \dfrac{\partial f}{\partial y} = \dfrac{2x^2y}{1+x^2y^2}\).
1.13. \(\displaystyle \quad \dfrac{\partial f}{\partial x} = 2xy \sin(z)\) and \(\displaystyle \dfrac{\partial f}{\partial y} = x^2 \sin(z)\) and \(\displaystyle \dfrac{\partial f}{\partial z} = x^2y \cos(z)\).
1.14. \(\displaystyle \quad \dfrac{\partial f}{\partial x} = (y+z)(2x+y+z)\) and \(\displaystyle \dfrac{\partial f}{\partial y} = (x+z)(x+2y+z)\) and \(\displaystyle \dfrac{\partial f}{\partial z} = (x+y)(x+y+2z)\).
1.15. \(\displaystyle \quad \dfrac{\partial f}{\partial x} = \dfrac{yz(y+z)}{(x+y+z)^2}\) and \(\displaystyle \dfrac{\partial f}{\partial y} = \dfrac{xz(x+z)}{(x+y+z)^2}\) and \(\displaystyle \dfrac{\partial f}{\partial z} = \dfrac{xy(x+y)}{(x+y+z)^2}\).
Q2
2.1. \(\displaystyle \quad \dfrac{\partial^2 f}{\partial x^2} = 2\) and \(\displaystyle \dfrac{\partial^2 f}{\partial y^2} = -2\) so \(\displaystyle \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} = 0\).
2.2. \(\displaystyle \quad \dfrac{\partial^2 f}{\partial x^2} = 0\) and \(\displaystyle \dfrac{\partial^2 f}{\partial y^2} = 0\) so \(\displaystyle \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} = 0\).
2.3. \(\displaystyle \quad \dfrac{\partial^2 f}{\partial x^2} = 6x\) and \(\displaystyle \dfrac{\partial^2 f}{\partial y^2} = -6x\) so \(\displaystyle \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} = 0\).
2.4. \(\displaystyle \quad \dfrac{\partial^2 f}{\partial x^2} = -\cos(x)\sinh(y)\) and \(\displaystyle \dfrac{\partial^2 f}{\partial y^2} = \cos(x)\sinh(y)\) so \(\displaystyle \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} = 0\).
2.5. \(\displaystyle \quad \dfrac{\partial^2 f}{\partial x^2} = e^x\sin(y)\) and \(\displaystyle \dfrac{\partial^2 f}{\partial y^2} = -e^x\sin(y)\) so \(\displaystyle \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} = 0\).
2.6. \(\displaystyle \quad \dfrac{\partial^2 f}{\partial x^2} = \dfrac{2xy}{(x^2+y^2)^2}\) and \(\displaystyle \dfrac{\partial^2 f}{\partial y^2} = -\dfrac{2xy}{(x^2+y^2)^2}\) so \(\displaystyle \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} = 0\).
2.7. \(\displaystyle \quad \dfrac{\partial^2 f}{\partial x^2} = \dfrac{2(y^2-x^2)}{(x^2+y^2)^2}\) and \(\displaystyle \dfrac{\partial^2 f}{\partial y^2} = \dfrac{2(x^2-y^2)}{(x^2+y^2)^2}\) so \(\displaystyle \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} = 0\).
Q3
3.1. \(\displaystyle \quad \dfrac{\partial^2 f}{\partial x \partial y} = 2x+2y\) and \(\displaystyle \dfrac{\partial^2 f}{\partial y \partial x} = 2x+2y\) so \(\displaystyle \dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{\partial^2 f}{\partial y \partial x}\).
3.2. \(\displaystyle \quad \dfrac{\partial^2 f}{\partial x \partial y} = -4x\sin(y)\) and \(\displaystyle \dfrac{\partial^2 f}{\partial y \partial x} = -4x\sin(y)\) so \(\displaystyle \dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{\partial^2 f}{\partial y \partial x}\).
3.3. \(\displaystyle \quad \dfrac{\partial^2 f}{\partial x \partial y} = 20(x+y)^3\) and \(\displaystyle \dfrac{\partial^2 f}{\partial y \partial x} = 20(x+y)^3\) so \(\displaystyle \dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{\partial^2 f}{\partial y \partial x}\).
3.4. \(\displaystyle \quad \dfrac{\partial^2 f}{\partial x \partial y} = -\dfrac{1}{(y+1)^2}\) and \(\displaystyle \dfrac{\partial^2 f}{\partial y \partial x} = -\dfrac{1}{(y+1)^2}\) so \(\displaystyle \dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{\partial^2 f}{\partial y \partial x}\).
3.5. \(\displaystyle \quad \dfrac{\partial^2 f}{\partial x \partial y} = -\dfrac{xy}{(x^2+y^2)^{3/2}}\) and \(\displaystyle \dfrac{\partial^2 f}{\partial y \partial x} = -\dfrac{xy}{(x^2+y^2)^{3/2}}\) so \(\displaystyle \dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{\partial^2 f}{\partial y \partial x}\).
3.6. \(\displaystyle \quad \dfrac{\partial^2 f}{\partial x \partial y} = 2x\cos(y) - 2y\sin(x)\) and \(\displaystyle \dfrac{\partial^2 f}{\partial y \partial x} = 2x\cos(y) - 2y\sin(x)\) so \(\displaystyle \dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{\partial^2 f}{\partial y \partial x}\).
3.7. \(\displaystyle \quad \dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{1-(xy)^2}{(1+(xy)^2)^2}\) and \(\displaystyle \dfrac{\partial^2 f}{\partial y \partial x} = \dfrac{1-(xy)^2}{(1+(xy)^2)^2}\) so \(\displaystyle \dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{\partial^2 f}{\partial y \partial x}\).
Version history and licensing
v1.0: initial version created 05/25 by Donald Campbell as part of a University of St Andrews VIP project.