1. Find the derivative of $y=(2x^{4}+5)(3x^{5}-8)$ directly using the product rule.\nProduct rule: $(fg)'=f'g+fg'$.\n$f=2x^{4}+5$, $g=3x^{5}-8$.\n$f'=8x^{3}$, $g'=15x^{4}$.\nSo, $y' = 8x^{3}(3x^{5}-8) + (2x^{4}+5)15x^{4}$.\nSimplify: $=24x^{8}-64x^{3} + 30x^{8} + 75x^{4} = (24x^{8} + 30x^{8}) + (-64x^{3} + 75x^{4}) = 54x^{8} + 75x^{4} - 64x^{3}$.\n\n2. Simplify $y=(2x^{4}+5)(3x^{5}-8)$ first, then differentiate.\nMultiply: $y=6x^{9} - 16x^{4} + 15x^{5} - 40$.\nCombine like terms: rearranged $y=6x^{9} + 15x^{5} - 16x^{4} - 40$.\nDerivative: $y' = 54x^{8} + 75x^{4} - 64x^{3}$ (same as above).\n\n3. Differentiate $y = (5x+8)^{2}$.\nUse chain rule: derivative of $u^{2}$ is $2u du/dx$, $u=5x+8$, $du/dx=5$.\n$y'=2(5x+8)(5) = 10(5x+8)$.\n\n4. Differentiate $y = \frac{3x^{8} - 4x^{7}}{4x^{3}}$.\nRewrite as $y = \frac{3x^{8}}{4x^{3}} - \frac{4x^{7}}{4x^{3}} = \frac{3}{4}x^{5} - x^{4}$.\nDerivative: $y' = \frac{3}{4}*5x^{4} - 4x^{3} = \frac{15}{4}x^{4} - 4x^{3}$.\n\n5. Differentiate $y = \frac{5x^{2} - 9x + 8}{x^{2} + 1}$.\nUse quotient rule: $(\frac{f}{g})' = \frac{f'g - fg'}{g^{2}}$, where $f=5x^{2} -9x +8$, $g=x^{2} + 1$.\n$f' = 10x - 9$, $g' = 2x$.\n$y' = \frac{(10x - 9)(x^{2}+1) - (5x^{2} - 9x + 8)(2x)}{(x^{2}+1)^{2}}$.\nExpand numerator:\n$(10x - 9)(x^{2} + 1) = 10x^{3} + 10x - 9x^{2} - 9$.\n$(5x^{2} - 9x + 8)(2x) = 10x^{3} - 18x^{2} + 16x$.\nSubtract: numerator = $10x^{3} + 10x - 9x^{2} - 9 - (10x^{3} - 18x^{2} + 16x) = (10x^{3}-10x^{3}) + (10x - 16x) + (-9x^{2} + 18x^{2}) - 9 = 0 - 6x + 9x^{2} - 9 = 9x^{2} - 6x - 9$.\nFinal derivative:\n$y' = \frac{9x^{2} - 6x - 9}{(x^{2} + 1)^{2}}$.\n\n6. Differentiate $y = \frac{1}{\sqrt{4x^{3} + 94}}$.\nRewrite $y = (4x^{3} + 94)^{-1/2}$.\nUse chain rule: $y' = -\frac{1}{2}(4x^{3}+94)^{-3/2} * 12x^{2} = -6x^{2} (4x^{3} + 94)^{-3/2}$.\n\n7. Use chain rule on $y = (4x^{5} - 1)^{7}$.\n$y' = 7(4x^{5} - 1)^{6} * 20x^{4} = 140 x^{4} (4x^{5} - 1)^{6}$.\n\n8. Use chain rule on $y = -3(x^{2} - 8x + 7)^{4}$.\n$y' = -3 * 4 (x^{2} - 8x + 7)^{3} * (2x - 8) = -12 (x^{2} - 8x + 7)^{3} (2x - 8)$.