Minta tolong ya! Deadlinenya jam 7 malam nih

1) diketahui f(x) = 2-2 sin (πx/2) dengan 0≤x≤π

a)) f'(x) = -2 ( π/2 ) cos ( πx/2)

= -π cos (πx/2)

b)) f'(x) = 0

– π cos (πx/2) = 0

cos (πx/2) = 0/-π

cos ( πx/2 ) = cos π/2

• πx/2 = π/2 + k . 2π

x = 0 + 4k

untuk k = 0 —> x = 0 ( memenuhi)

k = 1 —-> x = 4 ( memenuhi )

k = 2 —–> x = 8 ( memenuhi )

k = 3 ——> x = 12 ( memenuhi )

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k = 45 —–> x = 180 ( memenuhi )

jadi Nilai x yang memenuhi adalah

{ 0,4,8,12,….., 180}

2) f(x) =sin ^5x / cos ⁴x

a)) f'(x) =[ 5 cos ⁴ x ( cos⁴x ) ]- [ sin^5x ( -4sin ³x)] / (cos ⁴x)²

f'(x) = 5 cos ^8x + 4 sin^8x / ( cos⁴x)²

b)) f'(π/4) = a

5 cos^8 ( 45°) + 4 sin ^8 (45°) / {cos⁴ ( 45° ) }² =a

[ 5( 1/2 . √2 ) ^8 ] + [ 4 ( 1/2. √2)^8 ] / (1/2.√2)² =a

[5(1/16)] + [ 4(1/16)] / (1/2) = a

9/16 = a/2

16a = 18

a = 9/8