Answer: see boxed answers below
Step-by-step explanation:
(i) multiply the exponent to the coefficient then subtract 1 from the exponent.
[tex]y=\dfrac{3}{5x^3}+3x^4+2x^2-20\\\\\\\text{rewrite it as follows}: y=\dfrac{3}{5}x^{-3}+3x^4+2x^2-20x^0\\\\\\y'=(-3)\dfrac{3}{5}x^{-3-1}+(4)3x^{4-1}+(2)2x^{2-1}-(0)20x^{0-1}\\\\\\y'=-\dfrac{9}{5}x^{-4}+12x^3+4x^1-0\\\\\\y'=\large\boxed{-\dfrac{9}{5x^{4}}+12x^3+4x}[/tex]
(ii) Use the division formula: [tex]y = \dfrac{a}{b}\rightarrow \quad y'=\dfrac{ab'-a'b}{b^2}[/tex]
[tex]a=5x^3+1\qquad \qquad a'=15x^2\\b=3x^5+4x^2\qquad \quad b'=15x^4+8x\\\\\\y'=\dfrac{(15x^2)(3x^5+4x^2)-(5x^3+1)(15x^4+8x)}{(3x^5+4x^2)^2}\\\\\\.\quad =\dfrac{45x^7+60x^4-75x^7-55x^4-8x}{(3x^5+4x^2)^2}\\\\\\.\quad =\large\boxed{\dfrac{-35x^7+5x^4-8x}{(3x^5+4x^2)^2}}[/tex]
