Answers: Multivariate chain rule
These are the answers to Questions: Multivariate chain rule.
Please attempt the questions before reading these answers!
Q1
1.1. \(\displaystyle \quad \dfrac{\mathrm{d}z}{\mathrm{d}t} = 2 e^{2t} \sin(t) ( \cos(t) + \sin(t) )\).
1.2. \(\displaystyle \quad \dfrac{\mathrm{d}z}{\mathrm{d}t} = \dfrac{3}{t} - \tan(t)\).
1.3. \(\displaystyle \quad \dfrac{\mathrm{d}z}{\mathrm{d}t} = \dfrac{3}{2}\sqrt{t} + 6t(t^2+1)^2\).
1.4. \(\displaystyle \quad \dfrac{\mathrm{d}z}{\mathrm{d}t} = \exp(t\ln(t+1)) \left( \ln(t+1) + \dfrac{t}{t+1} \right) = ( t+1 ) ^t \left( \ln(t+1) + \dfrac{t}{t+1} \right)\).
1.5. \(\displaystyle \quad \dfrac{\mathrm{d}z}{\mathrm{d}t} = 2t\cos(t)\sec^2(t^2) - \sin(t)\tan(t^2)\).
1.6. \(\displaystyle \quad \dfrac{\mathrm{d}z}{\mathrm{d}t} = 15\cos(t) ( 2t - 1 + 25 \sin^2(t) ) + 8t - 4 + 30\sin(t)\).
1.7. \(\displaystyle \quad \dfrac{\mathrm{d}z}{\mathrm{d}t} = \dfrac{2t}{t-2} - \dfrac{t^2+1}{(t-2)^2} = \dfrac{t^2-4t-1}{(t-2)^2}\).
1.8. \(\displaystyle \quad \dfrac{\mathrm{d}z}{\mathrm{d}t} = 0\).
1.9. \(\displaystyle \quad \dfrac{\mathrm{d}z}{\mathrm{d}t} = t^2 e^t ( t^4 + 6t^3 + e^t (2t + 3) )\).
1.10. \(\displaystyle \quad \dfrac{\mathrm{d}z}{\mathrm{d}t} = \dfrac{2}{t} + te^{-t}(2-t)\).
1.11. \(\displaystyle \quad \dfrac{\mathrm{d}z}{\mathrm{d}t} = 4t ( 2\ln(t) + 1 )\).
1.12. \(\displaystyle \quad \dfrac{\mathrm{d}z}{\mathrm{d}t} = 3(t^3+1) ( 2t^2 \sin(3t) + (t^3+1)\cos(3t) )\).
1.13. \(\displaystyle \quad \dfrac{\mathrm{d}z}{\mathrm{d}t} = \dfrac{1}{t^2 + 1}\).
1.14. \(\displaystyle \quad \dfrac{\mathrm{d}z}{\mathrm{d}t} = \dfrac{\exp(\sqrt{t})}{t+2} + \dfrac{\ln(t+2) \exp(\sqrt{t})}{2\sqrt{t}}\).
Q2
2.1. \(\displaystyle \quad \dfrac{\partial z}{\partial s} = 2(s+t)(2s^2+st-t^2)\) and \(\displaystyle \dfrac{\partial z}{\partial t} = 2(s+t)(s^2-st-2t^2)\).
2.2. \(\displaystyle \quad \dfrac{\partial z}{\partial s} = 1\) and \(\displaystyle \dfrac{\partial z}{\partial t} = \dfrac{\cos(t)-\sin(t)}{\cos(t)+\sin(t)}\).
2.3. \(\displaystyle \quad \dfrac{\partial z}{\partial s} = 3t(s^2t^2-2s-t)\) and \(\displaystyle \dfrac{\partial z}{\partial t} = 3s(s^2t^2-s-2t)\).
2.4. \(\displaystyle \quad \dfrac{\partial z}{\partial s} = 2st \exp(s^2)\) and \(\displaystyle \dfrac{\partial z}{\partial t} = \exp(s^2)\).
2.5. \(\displaystyle \quad \dfrac{\partial z}{\partial s} = \sin(st) + t(s-t^2) \cos(st)\) and \(\displaystyle \dfrac{\partial z}{\partial t} = -2t \sin(st) + s(s-t^2) \cos(st)\).
2.6. \(\displaystyle \quad \dfrac{\partial z}{\partial s} = 2\sin(s)\cos(s)\cos(2t)\) and \(\displaystyle \dfrac{\partial z}{\partial t} = 2\sin(t)\cos(t)\cos(2s)\).
2.7. \(\displaystyle \quad \dfrac{\partial z}{\partial s} = 4s+2t\) and \(\displaystyle \dfrac{\partial z}{\partial t} = 2s\).
2.8. \(\displaystyle \quad \dfrac{\partial z}{\partial s} = \dfrac{1}{s+t} - \dfrac{1}{s}\) and \(\displaystyle \dfrac{\partial z}{\partial t} = \dfrac{1}{s+t} - \dfrac{1}{t}\).
2.9. \(\displaystyle \quad \dfrac{\partial z}{\partial s} = (2s+1)\sec^2(s^2+s+t^2-t)\) and \(\displaystyle \dfrac{\partial z}{\partial t} = (2t-1)\sec^2(s^2+s+t^2-t)\).
2.10. \(\displaystyle \quad \dfrac{\partial z}{\partial s} = -\dfrac{2t}{s^2+t^2}\) and \(\displaystyle \dfrac{\partial z}{\partial t} = \dfrac{2s}{s^2+t^2}\).
Q3
3.1. \(\displaystyle \quad \dfrac{\partial w}{\partial s} = 2s(t^2+2)\) and \(\displaystyle \dfrac{\partial w}{\partial t} = 2t(s^2+2)\).
3.2. \(\displaystyle \quad \dfrac{\partial w}{\partial s} = t(2s+t+u)\) and \(\displaystyle \dfrac{\partial w}{\partial t} = s(s+2t+u)+1\) and \(\displaystyle \dfrac{\partial w}{\partial u} = st+1\).
3.3. \(\displaystyle \quad \dfrac{\partial w}{\partial s} = 2st^2\cos(s^2t^2) - \sin(s+t)\) and \(\displaystyle \dfrac{\partial w}{\partial t} = 2s^{2}t\cos(s^2t^2) - \sin(s+t)\).
3.4. \(\displaystyle \quad \dfrac{\partial w}{\partial s} = 4s+4u\) and \(\displaystyle \dfrac{\partial w}{\partial t} = 4t\) and \(\displaystyle \dfrac{\partial w}{\partial u} = 4s+4u\).
Version history and licensing
v1.0: initial version created 05/25 by Donald Campbell as part of a University of St Andrews VIP project.