Matematika

Pertanyaan

tolong dibantu lagi ya kak no 24
tolong dibantu lagi ya kak no 24

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2 Jawaban

  • = {3(2x-1)-2(3x+1) /(2x-1)²} -sin {(3x+1)/(2x-1)}
    = {(6x-3-6x-2)/(2x-1)²} - sin {(3x+1)/(2x-1)}
    = {-5/(2x-1)²} - sin {(3x+1)/(2x-1)}
    = 5/(2x-1)² sin {(3x+1)(2x-1)}

  • [tex]f (x) = cos (\frac{3x + 1}{ 2x - 1} ) .... dengan , memisalkan... g (x) = \frac{3x + 1}{2x - 1} [/tex]

    [tex]g (x) = \frac{3x + 1}{2x - 1} ... dgn ,u = 3x + 1 ==\ \textgreater \ u' = 3 ; dan ,v = 2x - 1 ==\ \textgreater \ u' = 2[/tex]

    [tex]g' (x) = \frac{u'.v - u.v'}{ v^{2} } [/tex]
     
            [tex]= \frac{3 (2x - 1) - 2 (3x+ 1)}{(2x - 1)^{2} } [/tex]
      
            [tex]= \frac{6x - 3 - 6x - 2}{ (2x - 1)^{2} } [/tex]

    [tex]g' (x) = \frac{- 5}{ (2x - 1)^{2} } [/tex]

    soal dapat dituli

    [tex]f (x) = cos (\frac{3x + 1}{2x - 1} )[/tex]

    f (x) = cos  g (x)

    f' (x) = g' (x) .f' (cos g (x))

    [tex]f' (x) = \frac{- 5}{ (2x - 1)^{2} } ( - sin (g (x) )[/tex]

    [tex]f' (x) = \frac{5}{ (2x - 1)^{2} } sin (\frac{3x + 1}{2x - 1} ) ... jawaban : D[/tex]

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