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F(x) = ( -2x³ – 2x + 1) . ( 3x² – 8)³ , f'(-1) ….?

F(x) = ( -2x³ – 2x + 1) . ( 3x² – 8)³ , f'(-1) ….?

jawaban untuk pertanyaan diatas adalah – 1.626

Konsep
f(x) = u(x). v(x)
f'(x) = u'(x). v(x) + v'(x). u(x)

f(x) = ax^m
f'(x) = m. a. x^(m-1)

f(x) =c, dengan c adalah konstanta
f'(x) = 0

f(x) = (ax^m+b) ^n
f'(x) = (a. m. x^(m-1)). n.(ax^m + b) ^(n-1)

(x-y)³ = x³ – 3x²y + 3xy² – y³

Diketahui
f(x) = ( -2x³ – 2x + 1) . ( 3x² – 8)³
u(x) = – 2x³ – 2x + 1
u'(x) = – 6x² – 2

v(x) = (3x²-8)³
v(x) = (27x^6 – 3(3x²)²(8) + 3(3x²)(8)² – 8³)
v(x) = (27x^6 – 216x⁴ + 576x² – 512)

v'(x) = (6)27x^5 – (4)216x³ + (2)576x
v'(x) = 162x^5 – 864x³ + 1.152x

f'(x) = u'(x). v(x) + v'(x). u(x)
= (-6x²-2)(27x^6 – 216x⁴ + 576x² – 512)+ (162x^5-864x³ + 1.152x)(-2x³-2x+1)
= -162x^8 + 1.296x^6 – 3.456x⁴ + 3.072x²-54x^6 + 432x⁴ – 1.152x² + 1024 – 324x^8 – 324x^6 + 162x^5 + 1.728x^6+1.728x⁴-864x³ – 2.304x⁴ – 2.304x² + 1.152x
= 162x^8 + 2.646x^6+162x^5-3.600x⁴-864x³-384x²+1.152x

f'(-1)
=162x^8 + 2.646x^6+162x^5-3.600x⁴-864x³-384x²+1.152x
= 162(-1)^8 + 2.646(-1)^6 +162(-1)^5 – 3.600(-1)⁴ – 864(-1)³ – 384(-1)² +1.152(-1)
= 162 + 2.646 – 162 – 3.600 + 864 – 384 – 1.152
= – 1.626

Jadi jawabnya adalah -1.626

 

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