Tentukan turunan fungsi :f(x) =x²-4x+6/2x²+2x-3

Tentukan turunan fungsi :f(x) =x²-4x+6/2x²+2x-3

Jawaban:

Rumus turunan dengan fungsi pecahan

$Largeboxed{sf f(x) = frac{u}{v} rightarrow f'(x) = frac{u'v - uv'}{v^{2}}} \$

$\$

$sf : f(x) = frac{x {}^{2} - 4x + 6 : rightarrow u}{2x {}^{2} + 2x - 3 : rightarrow : v} \$

dengan

u = x² – 4x + 6
u' = 2x – 4
v = 2x² + 2x – 3
v' = 4x + 2

$\$

$sf f'(x) = frac{(2x - 4)(2x {}^{2} + 2x - 3) - ( {x}^{2} - 4x + 6)(4x + 2) }{(2x {}^{2} + 2x - 3) {}^{2} } \$

$sf f'(x) = frac{(4x {}^{3} + 4x {}^{2} - 6x - 8x {}^{2} - 8x + 12) - (4x {}^{3} + 2x {}^{2} - 16x {}^{2} - 8x + 24x + 12 }{(2x {}^{2} + 2x - 3 {}^{2} } \$

$sf f'(x) = frac{(4x {}^{3} - 4x {}^{2} - 14x + 12) - (4x {}^{3} - 14x {}^{2} + 16x {}^{2} + 12 )}{(2x {}^{2} + 2x - 3) {}^{2} } \$

$sf f'(x) = frac{ - 4x {}^{2} + 14x {}^{2} - 14x - 16x}{(2x {}^{2} + 2x - 3) {}^{2} )} \$

$sf f'(x) = frac{10 {}^{2} - 30x}{(2x {}^{2} + 2x - 3) {}^{2} } \ \$

Detail jawaban

♬ Mapel : Matematika

♬ Kelas : XI

♬ Materi : Bab 9 – Turunan fungsi aljabar

♬ Kode mapel : 2

♬ Kode kategorisasi : 11.2.9

♬ Kata kunci : Turunan pecahan

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