Answers: Introduction to differentiation and the derivative
These are the answers to Questions: Introduction to differentiation and the derivative.
Please attempt the questions before reading these answers!
1.1. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^3+5x-3\right)=3x^2+5.\)
1.2. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(5x\right)=5.\)
1.3. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(-5\sqrt{x}\right)=-5\cdot\frac{1}{2}x^{-1/2}=-\frac{5}{2\sqrt{x}}.\)
1.4. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(-\sin(x)\right)=-\cos(x).\)
1.5. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\cos x+5\right)=-\sin(x).\)
1.6. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(2\sqrt{x}\right)=2\cdot\frac{1}{2}x^{-1/2}=\frac{1}{\sqrt{x}}.\)
1.7. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(2\ln(2x)+x^5\right)=\frac{2}{x}+5x^4.\)
1.8. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\ln(5x)\right)=\frac{1}{5x}\cdot5=\frac{1}{x}.\)
1.9. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(e^{-x}\right)=e^{-x}\cdot(-1)=-e^{-x}.\)
1.10. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(23x+5\right)=23.\)
1.11. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(4x+100\right)=4.\)
1.12. For \(\displaystyle \sinh(5x)= \frac{e^{5x} - e^{-5x}}{2}\), it follows that \[\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\sinh(5x)\right)=5 \frac{e^{5x} +e^{-5x}}{2} = 5\cosh(5x)\] since \(\displaystyle\cosh(x) = \frac{e^x +e^{-x}}{2}\).
1.13. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\cos(3x)-\sin(2x)\right)=-3\sin(3x)-2\cos(2x).\)
1.14. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\ln(x)+\cos(x)+3x\right)=\frac{1}{x}-\sin(x)+3.\)
1.15. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\frac{2}{5}\sinh(x)+\frac{2}{13}\cosh(x)\right)=\frac{2}{5}\cosh(x)+\frac{2}{13}\sinh(x).\)
1.16. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(e^{5x}+x^2+3\right)=5e^{5x}+2x.\)
1.17. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\ln(x)+x^2\right)=\frac{1}{x}+2x.\)
1.18. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\ln(5x)-\ln(x)\right)=\frac{1}{x}-\frac{1}{x}=0.\)
1.19. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\cosh(x)-5x^7\right)= \sinh(x)-35x^6.\)
1.20. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\sqrt{3x^2}\right)=\sqrt{3}\)
1.21. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^3+3x-\sqrt{2x}\right)=3x^2+3-\frac{1}{\sqrt{2x}}.\)
Version history and licensing
v1.0: initial version created 05/25 by Sara Delgado Garcia as part of a University of St Andrews VIP project.