1. Misalkan f(x)= sin(sinx.cosx), maka f'(x)= . . . a. cos(sinx.cosx) b. sin (cos^2 x -sin^2 x) c. cos (sin x). cos x(cos x) d. cos 2x. cos (sinx.cosx) 2. jika

2 sin x cos x = sin 2x => sin x cos x = 1/2 sin 2x

1) f(x) = sin (sin x cos x) = sin (1/2 sin 2x)
f'(x) = cos (1/2 sin 2x) . 1/2 cos 2x . 2
= cos 2x . cos (1/2 sin 2x)
= cos 2x . cos (sin x cos x)

2) y - 2x + 5 = 0
=> y = 2x - 5 => m = 2
x = 2 => y = 2(2) - 5 => y = -1

(2, -1)
y = px^3 - qx^2 + 1
-1 = p(2)^3 - q(2)^2 + 1
-1 = 8p - 4q + 1
-8p + 4q = 2
-4p + 2q = 1

y = px^3 - qx^2 + 1
y' = 3px^2 - 2qx
m = y'
2 = 3p(2)^2 - 2q(2)
2 = 12p - 4q
1 = 6p - 2q

Eliminasi
-4p + 2q = 1
6p - 2q = 1
----------------- +
2p = 2 => p = 1

-4p + 2q = 1
-4(1) + 2q = 1
2q = 5
q = 5/2

2pq = 2(1)(5/2) = 5

1) misalkan U=sin x. cos x
Mk U' = cos²x - sin²x
f(x) = sin U
f'(x) = U' cos U = (cos²x-sin²x) cos(sin x. cos x) = cos 2x cos (sin x.cos x) (D)

2) y-2x+5 = 0 masukkan x=2 mk di dpt y = - 1
Masukkan (2,-1) ke y=px³-qx²+1 mk di dpt 8p-4q+1 = - 1 atau 8p-4q = - 2
y' = 3px²-2qx masukkan x=2 shg y'=12p-4q
y-2x+5=0 memiliki gradien = 2
Mk 12p-4q = 2
Eleminasi 8p-4q = - 2 dengan 12p-4q = 2 akan di dapat 4p = 4 atau p=1
p=1 —> 8-4q = - 2 —> q = 2½
2pq = 2.1.2½ = 5 (A)