Question

kuis turunan part 1 :

1. turunan dari :
[tex]f(x) = \frac{1}{5} x {}^{20} - 6 \: \sqrt[3]{x {}^{2} } [/tex]

2. Turunan dari :
[tex]f(x) = 25( \frac{2x {}^{3} }{6} + \frac{1}{ {5}^{5} \sqrt{x} } - x {}^{2} - \sqrt{99} ) {}^{} [/tex]


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Answer 1

Jawab:1)[tex]f(x) = \frac{1}{5}x^{20} - 6\sqrt[3]{x^2} = \frac{1}{5}x^{20} - 6x^{\frac{2}{3} }\\\\f'(x) = 4x^{19} - 4x^{-\frac{1}{3} }[/tex]2)[tex]f(x) = 25(\frac{2x^3}{6} + \frac{1}{5\sqrt[5]{x} } - x^2 - \sqrt{99})\\ = \frac{25}{3}x^3 + \frac{5}{\sqrt[5]{x} } - 25x^2 - 25\sqrt{99}\\ \\f'(x) = 25x^2 - \frac{1}{x^{\frac{6}{5} }} - 50x[/tex]

Answer 2

Jawab:1. [tex]f(x) = \frac{1}{5} x {}^{20} - 6\sqrt[3]{x {}^{2} }[/tex][tex]f(x) = \frac{1}{5} x {}^{20} - 6x^{\frac{2}{3} } \\[/tex][tex]f'(x) = \frac{1}{5} .20x^{20-1}-\frac{2}{3} .6x^{\frac{2}{3} -1} \\[/tex][tex]f'(x) = 4x^{19}-4x^{-\frac{1}{3} } \\[/tex][tex]f'(x) = 4x^{19}-4\frac{1}{x^{\frac{1}{3} } }[/tex][tex]f'(x) = 4x^{19}-4\frac{1}{\sqrt[3]{x} } } }[/tex]2.[tex]f(x)=25(\frac{2x^{3} }{6} +\frac{1}{5^{5} \sqrt{x} } -x^{2} -\sqrt{99})[/tex][tex]f(x)=25(\frac{1}{3}x^{3} +\frac{1}{5^{5} x^{\frac{1}{2} } } -x^{2} -99^{\frac{1}{2} } )[/tex][tex]f(x)=25(\frac{1}{3}x^{3} +\frac{1}{5^{5}}. \frac{1}{x^{\frac{1}{2} } } -x^{2} -99^{\frac{1}{2} } )[/tex][tex]f(x)=25(\frac{1}{3}x^{3} +\frac{1}{5^{5}}.x^{-\frac{1}{2} } -x^{2} -99^{\frac{1}{2} } )[/tex][tex]f'(x)=25(3.\frac{1}{3}x^{3-2} +(-\frac{1}{2}).\frac{1}{5^{5}} x^{-\frac{1}{2}-1 } -2x^{2-1} )[/tex][tex]f'(x)=25(x-\frac{1}{2.5^5} x^{-\frac{3}{2} } -2x )[/tex][tex]f'(x)=25(x-\frac{1}{2.5^5}. \frac{1}{x^{\frac{3}{2} }} -2x )[/tex][tex]f'(x)=25(x-\frac{1}{2.5^5}. \frac{1}{\sqrt{x^{3} }} -2x )[/tex][tex]f'(x)=25(x- \frac{1}{2.5^5\sqrt{x^{3} }} -2x )[/tex][tex]f'(x)=25x- \frac{1}{2.5^3\sqrt{x^{3} }} -50x[/tex][tex]f'(x)=25x- \frac{1}{250\sqrt{x^{3} }} -50x[/tex]Penjelasan:1.[tex]f(x) = g(x)-h(x)\\[/tex] [tex]f'(x) = g'(x)-h'(x)[/tex]2.[tex]f(x)=k.u(x)\\[/tex][tex]f'(x)=k.u'(x)[/tex]