Answer:
dy/dx = [tex]\frac{1}{(4x^{3}-7)}*[\frac{(3x^{5}+1)(12x^{2})-(4x^{3}-7)(15x^{4})}{(3x^{5}+1)}][/tex]
Step-by-step 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 = [tex]\frac{(3x^{5}+1)(12x^{2})-(4x^{3}-7)(15x^{4})}{(3x^{5}+1)^{2}}[/tex]
∵ The derivative of ㏑(f(x)) = 1/f(x) × f'(x)
∴ dy/dx = [tex]\frac{1}{\frac{(4x^{3}-7)}{(3x^{5}+1)}}*[\frac{(3x^{5}+1)(12x^{2})-(4x^{3}-7)(15x^{4}}{(3x^{5}+1)^{2}}][/tex]
- Simplify by cancel bracket (3x^5 + 1)from the 1st fraction with the
same bracket in the 2nd fraction
∴ dy/dx = [tex]\frac{1}{(4x^{3}-7)}*[\frac{(3x^{5}+1)(12x^{2})-(4x^{3}-7)(15x^{4})}{(3x^{5}+1)}][/tex]
