Jika f(x)=(2x+(1/(2x))^2 maka f''(1)=....

19/2

Pembahasan

Pertama, kita jabarkan fungsi f(x):
f(x) = (2x + (2x)^-1)^2
f(x) = (2x)^2 + 2 * (2x) * (2x)^-1 + ((2x)^-1)^2
f(x) = 4x^2 + 2 * (2x)^0 + (2x)^-2
f(x) = 4x^2 + 2 + (1/(4x^2))

Selanjutnya, kita cari turunan pertama f'(x):
f'(x) = d/dx (4x^2 + 2 + (1/4)x^-2)
f'(x) = 8x + 0 + (1/4) * (-2)x^-3
f'(x) = 8x - (1/2)x^-3
f'(x) = 8x - 2/(4x^3)
f'(x) = 8x - 1/(2x^3)

Kemudian, kita cari turunan kedua f''(x):
f''(x) = d/dx (8x - (1/2)x^-3)
f''(x) = 8 - (1/2) * (-3)x^-4
f''(x) = 8 + (3/2)x^-4
f''(x) = 8 + 3/(2x^4)

Terakhir, kita substitusikan x=1 ke dalam f''(x):
f''(1) = 8 + 3/(2*(1)^4)
f''(1) = 8 + 3/2
f''(1) = 16/2 + 3/2
f''(1) = 19/2