Answers: The product rule
These are the answers to Questions: The product rule.
Please attempt the questions before reading these answers!
1.1. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(xe^x\right)=e^x + xe^x.\)
1.2. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^2e^{2x}\right)=2xe^{2x}+2x^2e^{2x}.\)
1.3. As \(\tan(x) = \sin(x)/cos(x)\), the function becomes \(5x^3\sin(x)\) and so \[\dfrac{\mathrm{d}}{\mathrm{d}x}\left(5x^3\tan(x)\cos(x)\right)= \dfrac{\mathrm{d}}{\mathrm{d}x}\left(5x^3\sin(x)\right) = 15x^2\sin(x) + 5x^3\cos(x).\]
1.4. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(x\ln(x)\right)=\ln(x)+1.\)
1.5. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left((x^3+x^2-5)(x+1)\right)=(3x^2+2x)(x+1)+(x^3+x^2-5).\)
1.6. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left((13x^2 + 5x +2)(x^3 + 2)\right)=(26x+5)(x^3 + 2) + 3x^2(13x^2 + 5x + 2).\)
1.7. Take the \(x\) inside the first bracket so the function becomes \((5x^3 + 3x^2 + 2x)(x^2 +x +1)\). Then \[ \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d}x}\left(x(5x^2 + 3x + 2)(x^2 +x +1)\right)&= \dfrac{\mathrm{d}}{\mathrm{d}x}\left((5x^3 + 3x^2 + 2x)(x^2 +x +1)\right)\\[1em] &= (10x^2+6x+2)(x^2 + x + 1) + (5x^3 + 3x^2 + 2x)(2x + 1). \end{aligned} \]
1.8. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left((10x^2 + 21)\cos(x)\right)=20x\cos(x) - (10x^2+21)\sin(x).\)
1.9. Using the definitions of \(\cosh(2x)\) and \(\sinh(3x)\):
\[ \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\cosh(2x)\sinh(3x)\right)&=\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\left(\frac{e^{2x}+e^{-2x}}{2}\right)\left(\frac{e^{3x} - e^{-3x}}{2}\right)\right)\\[1em] &= \left(\frac{e^{2x}+e^{-2x}}{2}\right)\left(\frac{3e^{3x} + 3e^{-3x}}{2}\right) + \left(\frac{2e^{2x}-2e^{-2x}}{2}\right)\left(\frac{e^{3x} - e^{-3x}}{2}\right)\\[1em] &= 3\cosh(2x)\cosh(3x) + 2\sinh(2x)\sinh(3x) \end{aligned} \]
1.10. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left((x^2 +3)\ln(x)\right)=2x\ln(x) + \frac{x^2 + 3}{x}.\)
1.11. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\sin(x)\sqrt{x}\right)=\cos(x)\sqrt{x} + \frac{\sin(x)}{2\sqrt{x}}.\)
1.12. Since \(\displaystyle \cosh(x)= \frac{e^{x} + e^{-x}}{2}\), it follows that \[ \begin{aligned}\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\cosh(x)\ln(x)\right)&=\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\left(\frac{e^{x}+e^{-x}}{2}\right)\ln(x)\right)\\[1em] &= \left(\frac{e^{x}-e^{-x}}{2}\right)\ln(x) + \left(\frac{e^{x}+e^{-x}}{2}\right)\frac{1}{x}\\[1em] &=\sinh(x)\ln(x) + \frac{\cosh(x)}{x} \end{aligned}\] since \(\displaystyle\sinh(x) = \frac{e^x -e^{-x}}{2}\).
1.13. Factorize to get \(x^2(\sqrt{x}+\cos(x))\), then \[ \begin{aligned}\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^2\sqrt{x}+x^2\cos(x)\right)&=\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^2(\sqrt{x}+\cos(x))\right)\\[1em] &= 2x(\sqrt{x}+\cos(x)) + x^2\left(\frac{1}{2\sqrt{x}} - \sin(x)\right).\end{aligned}\]
1.14. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(e^{-5x}(x^3+5)\right)=3x^2e^{-5x} - 5e^{-5x}(x^3+5).\)
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. Using the product rule twice here: \[\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\ln(x) \ln (3x) \ln (100x)\right)=\frac{\ln(3x)\ln(100x) + \ln(x)\ln(100x) + \ln(x)\ln(3x)}{x}.\]
1.17. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left((x^2 + 5x + 2)\sin(x)\right)=(2x+5)\sin(x) + (x^2 + 5x + 2)\cos(x).\)
1.18. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left(-\ln(x)\ln(3x)\right)= -\frac{1}{x}(\ln(3x)+\ln(x)).\)
1.19. Using the product rule twice: \[ \begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d}x}&\left((x^5 + 3)(x^2 + 3x)(x^7 + x^4)\right)\\ &= 5x^4(x^2 +3x)(x^7 + x^4) + (x^5 + 3)(2x + 3)(x^7 + x^4) + (x^5 +3)(x^2 + 3x)(7x^6 + 4x^3). \end{aligned} \]
1.20. \(\displaystyle \quad \dfrac{\mathrm{d}}{\mathrm{d}x}\left((\sin(x) + 3x)e^{-x}\right)= (\cos(x) + 3)e^{-x} - (\sin(x) + 3x)e^{-x}.\)
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.