Question

Find the derivative in the form dy/dx

Answer 1

dy/dx =
`(1)/((4x^(3)-7))*[((3x^(5)+1)(12x^(2))-(4x^(3)-7)(15x^(4)))/((3x^(5)+1))]`

Explanation:

* Lets revise some rules for the derivative

- The derivative of ㏑(f(x)) = 1/f(x) × f'(x)

- The derivative of u/v = (vu'-uv')/v²

- The derivative of the constant is 0

* Lets solve the problem

∵ y = ㏑[(4x³ - 7)/(3x^5 + 1)]

- Let u = 4x³ - 7 and v = 3x^5 + 1

∵ u = 4x³ - 7

∴ u' = 4(3)x^(3-1) - 0 = 12x²

∵ v = 3x^5 + 1

∴ v' = 3(5)x^(5-1) + 0 = 15x^4

∵ The derivative of u/v = (vu' - uv')/v²

∴ The derivative of u/v =
`((3x^(5)+1)(12x^(2))-(4x^(3)-7)(15x^(4)))/((3x^(5)+1)^(2))`

∵ The derivative of ㏑(f(x)) = 1/f(x) × f'(x)

∴ dy/dx =
`(1)/(((4x^(3)-7))/((3x^(5)+1)))*[((3x^(5)+1)(12x^(2))-(4x^(3)-7)(15x^(4))/((3x^(5)+1)^(2))]`

- Simplify by cancel bracket (3x^5 + 1)from the 1st fraction with the

same bracket in the 2nd fraction

∴ dy/dx =
`(1)/((4x^(3)-7))*[((3x^(5)+1)(12x^(2))-(4x^(3)-7)(15x^(4)))/((3x^(5)+1))]`