Question

tolong dgn cara jgn asal

ini ttg turunan fungsi aljabar

SMA MTK yg ga bisa ga usah jwb gpp g mksa ​

Answer 1

Penjelasan dengan langkah-langkah:

no. 17

Ingat !

f(x) = (ax + b)/(cx + d)

f'(x) = (ad - bc)/(cx + d)²

maka :

f(x) = (x - 36)/(x + 6)

f'(x) = (1 . 6 - (-36) . 1)/(x + 6)²

f'(x) = (6 + 36)/(x + 6)²

f'(x) = 42/(x + 6)²

Tentukan nilai dari f(0) + 6f'(0) !

Jawab :

f(0) = (0 - 36)/(0 + 6) = -36/6 = -6

f'(0) = 42/(0 + 6)² = 42/36 = 7/6

Maka, nilai dari f(0) + 6f'(0) adalah 1

f(0) + 6f'(0)

= -6 + 6(7/6)

= -6 + 7

= 1

no. 18

f(x) = (√(x + 1))/x

• u = (x + 1)^½ dan u' = ½ / √(x + 1)

• v = x dan v' = 1

maka :

f(x) = u/v

f'(x) = (u' . v - u . v')/v²

f'(x) = ((½/√(x + 1) . x) - (√(x + 1) . 1)/x²

f'(x) = (x / 2√(x + 1) - √(x + 1))/x²

f'(3) = (3 / 2√(3 + 1) - √(3 + 1))/3²

f'(3) = (¾ - 2)/9

f'(3) = -5/4 × 1/9

f'(3) = -5/36

Semoga Bermanfaat