Question

How to differentiate functions​

Answer 1

Answer: see boxed answers below

Explanation:

(i) multiply the exponent to the coefficient then subtract 1 from the exponent.

`y=(3)/(5x^3)+3x^4+2x^2-20\\\\\\\text{rewrite it as follows}: y=(3)/(5)x^(-3)+3x^4+2x^2-20x^0\\\\\\y'=(-3)(3)/(5)x^(-3-1)+(4)3x^(4-1)+(2)2x^(2-1)-(0)20x^(0-1)\\\\\\y'=-(9)/(5)x^(-4)+12x^3+4x^1-0\\\\\\y'=\large\boxed{-(9)/(5x^(4))+12x^3+4x}`

(ii) Use the division formula:
`y = (a)/(b)\rightarrow \quad y'=(ab'-a'b)/(b^2)`

`a=5x^3+1\qquad \qquad a'=15x^2\\b=3x^5+4x^2\qquad \quad b'=15x^4+8x\\\\\\y'=((15x^2)(3x^5+4x^2)-(5x^3+1)(15x^4+8x))/((3x^5+4x^2)^2)\\\\\\.\quad =(45x^7+60x^4-75x^7-55x^4-8x)/((3x^5+4x^2)^2)\\\\\\.\quad =\large\boxed{(-35x^7+5x^4-8x)/((3x^5+4x^2)^2)}`