Tolong bantu jawab! Tentukan dy/dx dari 1. y= log (x²-7)³ 2. y= In (x+2/x-3) 3. y= (In 4x³)² Mohon bantu jawab

jawab

Turunan dy/dx dari 

1.
y = log (x² -7)³
y =  3. log (x² - 7)

u  = x² - 7 
du/dx = 2x

y = 3 log u
dy/du  = 3/u.ln 10

dy/dx = dy/du . du/dx
y ' = 3/(u lln 10) . (2x)
y' = 6x / (x² -7),ln10

2, y = ln {(x + 2)/(x -3)}
u = (x+2)/(x-3)
du/dx = -1/(x-3)²

y = ln u
dy/du = 1/ u

dy/dx  = dy/du. du/dx
y' = 1/u. (-1)/(x-3)²
y' =  - (x-3)/(x +2)(x - 3)²
y' = - 1/(x+2)(x-3)

3)
y = (ln  4x³)²
t = ln 4x³
dt/dx= (12 x²)/(4x³) = 3/x

y = t²
dy/dt = 2t

dy/dx = dy/dt. dt/dx
y' = (2t). (3/x)
y' = 6t/x
y' = 6 ln 4x³  / x