Find the first derivative

f(x)=(x^(2)-x)/((2x+4)^7)

f(x)=((√(x)-x^2)^9) √(x^(2)+√(2))

For each expression, set both sides by logs. Then, differentiate w/respect to x, noting that d/dx ln(f(x)) = f'(x)/f(x)!

1. lnf(x) = ln(x² - x) - 7ln(2x + 4)
f'(x)/f(x) = (2x - 1)/(x² - x) - 7(2)/(2x + 4)
f'(x) = f(x)((2x - 1)/(x² - x) - 14/(2x + 4)) [Multiply both sides by f(x)]
f'(x) = (x² - x)/(2x + 4)^7 * ((2x - 1)/(x² - x) - 14/(2x + 4)) [Substitute f(x) back with original function]

2. ln(f(x)) = 9ln(√x - x²) + ½ln(x² + √(2))
f'(x)/f(x) = 9(1/(2√x) - 2x)/(√x - x²) * 2√x/(2√x) + ½ * 2x/(x² + √(2))
f'(x)/f(x) = 9(1 - 4x^(3/2))/(2√(x)(√x - x²)) + x/(x² + √(2))
f'(x) = f(x) * (9(1 - 4x^(3/2))/(2√(x)(√x - x²)) + x/(x² + √(2)))
f'(x) = ((√(x) - x²)^9)√(x² + √(2)) * (9(1 - 4x^(3/2))/(2√(x)(√x - x²)) + x/(x² + √(2)))

I hope this helps!