Dengan Jalannya Ya Kak ;)
f(x) = √(2x³ - 1)
Konsep turunan :
[tex] = \lim \limits_{h \to0} \frac{f(x + h) - f(x)}{h} [/tex]
[tex] = \lim \limits_{h \to0} \frac{ \sqrt{2 {(x + h)}^{3} - 1 } - \sqrt{2 {x}^{3} - 1} }{h} [/tex]
[tex] = \lim \limits_{h \to0} \frac{ \sqrt{2( {x}^{3} + 3 {x}^{2}h + 3x {h}^{2} + {h}^{3}) - 1 } - \sqrt{2 {x}^{3} - 1} }{h} [/tex]
[tex] = \lim \limits_{h \to0} \frac{ \sqrt{2 {x}^{3} + 6 {x}^{2} h + 6x {h}^{2} + 2 {h}^{3} - 1} - \sqrt{2 {x}^{3} - 1 } }{h} [/tex]
[tex] = \lim \limits_{h \to0} \frac{ \sqrt{2 {x}^{3} + 6 {x}^{2} h + 6x {h}^{2} + 2 {h}^{3} - 1} - \sqrt{2 {x}^{3} - 1 } }{h} \times \frac{ \sqrt{2 {x}^{3} + 6 {x}^{2} h + 6x {h}^{2} + 2 {h}^{3} - 1 } + \sqrt{2 {x}^{3} - 1 } }{\sqrt{2 {x}^{3} + 6 {x}^{2} h + 6x {h}^{2} + 2 {h}^{3} - 1 } + \sqrt{2 {x}^{3} - 1 } } [/tex]
[tex] = \lim \limits_{h \to0} \frac{ (\sqrt{2 {x}^{3} + 6 {x}^{2} h + 6x {h}^{2} + 2 {h}^{3} - 1) } {}^{2} - (\sqrt{2 {x}^{3} - 1 }) {}^{2} }{h(\sqrt{2 {x}^{3} + 6 {x}^{2} h + 6x {h}^{2} + 2 {h}^{3} - 1 } + \sqrt{2 {x}^{3} - 1 })} [/tex]
[tex] = \lim \limits_{h \to0} \frac{ ( \cancel{2 {x}^{3}} + 6 {x}^{2} h + 6x {h}^{2} + 2 {h}^{3} - \cancel1) - ( \cancel{2 {x}^{3}} - \cancel1 ) }{h(\sqrt{2 {x}^{3} + 6 {x}^{2} h + 6x {h}^{2} + 2 {h}^{3} - 1 } + \sqrt{2 {x}^{3} - 1 })} [/tex]
[tex] = \lim \limits_{h \to0} \frac{ 6 {x}^{2} h + 6x {h}^{2} + 2 {h}^{3} }{h(\sqrt{2 {x}^{3} + 6 {x}^{2} h + 6x {h}^{2} + 2 {h}^{3} - 1 } + \sqrt{2 {x}^{3} - 1 })} [/tex]
[tex] = \lim \limits_{h \to0} \frac{ \cancel h( 6 {x}^{2} + 6x {h} + 2 {h}^{2} ) }{ \cancel h(\sqrt{2 {x}^{3} + 6 {x}^{2} h + 6x {h}^{2} + 2 {h}^{3} - 1 } + \sqrt{2 {x}^{3} - 1 })} [/tex]
[tex] = \lim \limits_{h \to0} \frac{ 6 {x}^{2} + 6x {h} + 2 {h}^{2} }{ \sqrt{2 {x}^{3} + 6 {x}^{2} h + 6x {h}^{2} + 2 {h}^{3} - 1 } + \sqrt{2 {x}^{3} - 1 }} [/tex]
[tex]= \frac{ 6 {x}^{2} + 6x {.0} + 2 {.0}^{2} }{ \sqrt{2 {x}^{3} + 6 {x}^{2} .0 + 6x {.0}^{2} + 2 {.0}^{3} - 1 } + \sqrt{2 {x}^{3} - 1 }} [/tex]
[tex] = \frac{ 6 {x}^{2} + 0 + 0 }{ \sqrt{2 {x}^{3} + 0 + 0 + 0 - 1 } + \sqrt{2 {x}^{3} - 1 }} [/tex]
[tex] = \frac{\cancel6 {x}^{2} }{\cancel2(\sqrt{2 {x}^{3} - 1} )} [/tex]
[tex] = \frac{3 {x}^{2} }{ \sqrt{2{x}^{3} - 1}}[/tex]
[tex] \text{turunan }\: f(x) = \frac{3{x}^{2} }{ \sqrt{2 {x}^{3} - 1} } [/tex]
Pembahasan :
Turunan dari f(g(x)) =
f'(g(x)) × g'(x)
soal:
f(x) = √(2x³-1)
= (2x³ - 1)¹´²
g(x) = (2x³-1)
g'(x) = 6x²
f(g(x)) = (2x³-1)¹´²
f'(x) = ½(2x³-1)⁻¹´² × 6x²
= 3x²/√(2x³-1)
[answer.2.content]
