Answer to Question #205822 in Calculus for Valerie Khumalo

Question

Use appropriate differentiation techniques to determine the first derivatives

1. $y=(3√x -2xe^x)/x$
2. $y=cos(√sin(tanπx))$
3. $y=(tanx-1)/secx$

Answer 1

$(1)\\ {}{{}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[\dfrac{3\sqrt{x}-2x\mathrm{e}^x}{x}\right]}}\\ =\dfrac{{}{{}{\left(3\cdot{}{{}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[\sqrt{x}\right]}}-2\cdot{}{{}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[x\mathrm{e}^x\right]}}\right)}}x-\left(3\sqrt{x}-2x\mathrm{e}^x\right)\cdot{}{{0}{1}}}{x^2}\\ =\dfrac{2x\mathrm{e}^x+\left(3\cdot{}{{1}{\frac{1}{2}}}{}{{2}{x^{\frac{1}{2}-1}}}-2\left({}{{}{{}{{}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[x\right]}}\cdot\mathrm{e}^x}}+{}{{}{x\cdot{}{{}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[\mathrm{e}^x\right]}}}}\right)\right)x-3\sqrt{x}}{x^2}\\ =\dfrac{2x\mathrm{e}^x+\left(\frac{3}{2\sqrt{x}}-2\left({}{{}{}}\mathrm{e}^x+x{}{{}{\mathrm{e}^x}}\right)\right)x-3\sqrt{x}}{x^2}\\ =\dfrac{x\left(\frac{3}{2\sqrt{x}}-2\left(x\mathrm{e}^x+\mathrm{e}^x\right)\right)+2x\mathrm{e}^x-3\sqrt{x}}{x^2}\\ Simplify/rewrite:\\ -\dfrac{4x^\frac{5}{2}\mathrm{e}^x+3x}{2x^\frac{5}{2}}\\ ----------------------\\ (2)\\ {}{{}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[\cos\left(\sqrt{\sin\left(\tan\left({\pi}x\right)\right)}\right)\right]}}\\ =-{}{{}{\dfrac{1}{2}}}{}{{}{\sin^{\frac{1}{2}-1}\left(\tan\left({\pi}x\right)\right)}}\cdot{}{{}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[\sin\left(\tan\left({\pi}x\right)\right)\right]}}\cdot\sin\left(\sqrt{\sin\left(\tan\left({\pi}x\right)\right)}\right)\\ =-\dfrac{{}{{}{\cos\left(\tan\left({\pi}x\right)\right)}}\cdot{}{{}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[\tan\left({\pi}x\right)\right]}}\cdot\sin\left(\sqrt{\sin\left(\tan\left({\pi}x\right)\right)}\right)}{2\sqrt{\sin\left(\tan\left({\pi}x\right)\right)}}\\ =-\dfrac{\cos\left(\tan\left({\pi}x\right)\right){}{{}{\sec^2\left({\pi}x\right)}}\cdot{}{{}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[{\pi}x\right]}}\cdot\sin\left(\sqrt{\sin\left(\tan\left({\pi}x\right)\right)}\right)}{2\sqrt{\sin\left(\tan\left({\pi}x\right)\right)}}\\ =-\dfrac{\cos\left(\tan\left({\pi}x\right)\right)\sec^2\left({\pi}x\right)\cdot{}{{1}{{\pi}\cdot{}{{}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[x\right]}}}}\cdot\sin\left(\sqrt{\sin\left(\tan\left({\pi}x\right)\right)}\right)}{2\sqrt{\sin\left(\tan\left({\pi}x\right)\right)}}\\ =-\dfrac{{\pi}\cos\left(\tan\left({\pi}x\right)\right)\sec^2\left({\pi}x\right)\cdot{}{{}{1}}\sin\left(\sqrt{\sin\left(\tan\left({\pi}x\right)\right)}\right)}{2\sqrt{\sin\left(\tan\left({\pi}x\right)\right)}}\\ =-\dfrac{{\pi}\sec^2\left({\pi}x\right)\cos\left(\tan\left({\pi}x\right)\right)\sin\left(\sqrt{\sin\left(\tan\left({\pi}x\right)\right)}\right)}{2\sqrt{\sin\left(\tan\left({\pi}x\right)\right)}}\\ -------------------------\\ (3)\\ {}{{}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[\dfrac{\tan\left(x\right)-1}{\sec\left(x\right)}\right]}}\\ =\dfrac{{}{{}{\left({}{{}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[\tan\left(x\right)\right]}}+{}{{}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[-1\right]}}\right)}}\sec\left(x\right)-\left(\tan\left(x\right)-1\right){}{{}{\sec\left(x\right)}}{}{{1}{\tan\left(x\right)}}}{\sec^2\left(x\right)}\\ =\dfrac{\left({}{{}{\sec^2\left(x\right)}}+{}{{}{0}}\right)\sec\left(x\right)-\left(\tan\left(x\right)-1\right)\sec\left(x\right)\tan\left(x\right)}{\sec^2\left(x\right)}\\ =\dfrac{\sec^3\left(x\right)-\sec\left(x\right)\left(\tan\left(x\right)-1\right)\tan\left(x\right)}{\sec^2\left(x\right)}\\ Rewrite/simplify:\\ =\sec\left(x\right)-\dfrac{\left(\tan\left(x\right)-1\right)\tan\left(x\right)}{\sec\left(x\right)}\\$