Matematika Sekolah Menengah Atas Quiz[tex] \: [/tex][tex] \rm \: f(x) = \sqrt[3]{2x + 3} [/tex]Maka:[tex] \rm \: f'(2) = \cdots[/tex]​

owenpriceulv – March 08, 2023

Quiz
[tex] \: [/tex]
[tex] \rm \: f(x) = \sqrt[3]{2x + 3} [/tex]

Maka:

[tex] \rm \: f'(2) = \cdots[/tex]
​

Jawaban:

Pembahasan

Turunan fungsi

[tex]f(x) = \sqrt[3]{2x + 3} [/tex]

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

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

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

[tex]f'(x) = \frac{2}{3} .(2x + 3) {}^{ - \frac{2}{3} } \\ [/tex]

[tex]f'(x) = \frac{2}{3.(2x + 3) {}^{ \frac{2}{3} } } \\ [/tex]

• Menentukan nilai dari f'(2)

[tex]f'(x) = \frac{2}{3.(2x + 3) {}^{ \frac{2}{3} } } \\ [/tex]

[tex]f'(2) = \frac{2}{3.(2(2) + 3)) {}^{ \frac{2}{3} } } \\ [/tex]

[tex]f'(2) = \frac{2}{3.(4 + 3) {}^{ \frac{2}{3} } } \\ [/tex]

[tex]f'(2) = \frac{2}{3 \: . \: {7}^{ \frac{2}{3} } } [/tex]

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