turunan fungsi menggunakan aturan rantai : 1. f(t) = √sin ([tex] \frac{1-t}{1+t} [/tex]) 2. y = f(x) = sin (cos√x) pakai cara yaaa +50 point

1) v = (1 - t)/(1 + t) = a/b
a = 1 - t => a' = -1
b = 1 + t => b' = 1
dv/dt = (a' b - b' a)/b^2 = (-1 (1 + t) - 1(1 - t))/(1 + t)^2
dv/dt = (-1 - t - 1 + t)/(1 + t)^2 = -2/(1 + t)^2

u = sin v
du/dv = cos v = cos (1 - t)/(1 + t)

f(t) = √u = u^1/2
df(t)/du = 1/2 u^-1/2 = 1/2 . 1/u^1/2 = 1/2 . 1/√u = 1/(2√u)
df(t)/du = 1/(2√(sin v)) = 1/(2√(sin (1 - t)/(1 + t))

f(t) = √(sin (1 - t)/(1 + t))
df(t)/dt = df(t)/du . du/dv . dv/dt
= 1/(2√(sin (1 - t)/(1 + t)) . cos (1 - t)/(1 + t) . -2/(1 + t)^2
= [-cos (1 - t)/(1 + t)] / (1 + t)^2√(sin (1 - t)/(1 + t))

2) y = sin (cos √x)
v = √x = x^1/2
dv/dx = 1/2 x^-1/2 = 1/2 . 1/x^1/2 = 1/2 . 1/√x = 1/(2√x)

u = cos v
du/dv = - sin v = - sin √x

y = sin u
dy/du = cos u = cos (cos v) = cos (cos √x)

dy/dx = dy/du . du/dv . dv/dx
= cos (cos √x) . - sin √x . 1/(2√x)
= -1/(2√x) . sin √x . cos (cos √x)