tolong,,,,,,,,,,,, bantuin yaa. makasih
Matematika
asyastie2
Pertanyaan
tolong,,,,,,,,,,,, bantuin yaa.
makasih
makasih
1 Jawaban
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1. Jawaban hendrisyafa
1) [tex]f'(x) = \frac{1}{3}x^{ \frac{1}{3}- \frac{3}{3} } = \frac{1}{3} x^{- \frac{2}{3} }= \frac{1}{3 \sqrt[3]{x^{2}} } [/tex]
2) [tex]f(x) = \frac{1}{x^{ \frac{1}{4} }} = x^{-\frac{1}{4} }[/tex]
[tex]f'(x)=- \frac{1}{4}x^{(- \frac{1}{4}+ \frac{4}{4}) } =- \frac{1}{4} x^{ \frac{3}{4} }=- \frac{1}{4} \sqrt[4]{x^{3}} [/tex]
3) [tex]f(x)= \frac{x^{4}}{2} [/tex]
[tex]f'(x)= \frac{4x^{4-1}}{2} =2x^{3}[/tex]
4) [tex]f(x) =8x^{ \frac{2}{3} }[/tex]
[tex]f'(x) =8. \frac{2}{3} x^{ (\frac{2}{3}- \frac{3}{3}) }= \frac{16}{3} x^{- \frac{1}{3} }= \frac{16}{ \sqrt[3]{x} }[/tex]
5) [tex]f(x) = 4x^{ \frac{1}{3} }[/tex]
[tex]f'(x)= 4. \frac{1}{3}x^{( \frac{1}{3}- \frac{3}{3} ) } = \frac{4x^{- \frac{2}{3} }}{3}= \frac{4}{3\sqrt[3]{x^{2}} } [/tex]
6) [tex]f'(x) = 4[/tex]
7) [tex]f'(x) = -4[/tex]
8) [tex]f(x) = x^{2} -3x[/tex]
[tex]f'(x) =2x-3[/tex]
9) [tex]f(x) =2 x^{2} +8x[/tex]
[tex]f'(x) = 4x+8[/tex]
10) [tex]f(x) = x^{2} -6x+9[/tex]
[tex]f'(x) =2x-6[/tex]
11) [tex]f'(x) =5x^{4}+9x^{2}-1[/tex]
12) [tex]f(x) =x^{ \frac{1}{2} }+x^{ \frac{1}{3} }[/tex]
[tex]f'(x) = \frac{1}{2} x^{- \frac{1}{2} }+ \frac{1}{3} x^{- \frac{2}{3} }= \frac{1}{2 \sqrt{x} } + \frac{1}{3 \sqrt[3]{x^{2}} }[/tex]