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Dikelahui f (0) = 1 dan f '(0) = 2. Jika g (x) = , maka g’(0) = ....

Pertanyaan

Dikelahui f (0) = 1 dan f '(0) = 2. Jika g (x) = begin mathsize 14px style 1 over open parentheses 2 f open parentheses x close parentheses minus 1 close parentheses cubed end style, maka g’(0) = ....

  1. -12

  2. -6

  3. 6

  4. 8

  5. 12

Pembahasan Soal:

g (x) = begin mathsize 14px style 1 over open parentheses 2 f open parentheses x close parentheses minus 1 close parentheses cubed end style

g (x) = (2f (x) - 1)-3

g' (x) = -3 ((2f (x) - 1)-4) . 2f ' (x)
g' (0) = -3 ((2f (0) - 1)-4) . 2f ' (0)
g' (0) = -3 ((2 (1) - 1)-4) . 2(2)
g' (0) = -3 (1)(4)
g' (0) = -12

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Desia

Mahasiswa/Alumni UIN Sunan Gunung Djati Bandung

Terakhir diupdate 17 Desember 2020

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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table attributes columnalign right center left columnspacing 0px end attributes row y equals cell square root of u end cell row y equals cell u to the power of 1 half end exponent end cell row cell fraction numerator straight d y over denominator straight d x end fraction end cell equals cell fraction numerator straight d y over denominator straight d u end fraction times fraction numerator straight d u over denominator straight d x end fraction end cell row blank equals cell fraction numerator straight d left parenthesis u to the power of 1 half end exponent right parenthesis over denominator straight d u end fraction times fraction numerator straight d left parenthesis 3 minus 4 x right parenthesis over denominator straight d x end fraction end cell row blank equals cell 1 half u to the power of negative 1 half end exponent times negative 4 end cell row blank equals cell negative 2 u to the power of negative 1 half end exponent end cell row blank equals cell fraction numerator negative 2 over denominator square root of u end fraction end cell row blank equals cell fraction numerator negative 2 over denominator square root of 3 minus 4 x end root end fraction end cell end table

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Tentukan turunan pertama fungsi-fungsi berikut. b. g(x)=(8x3−x2+3x−5)−5

Pembahasan Soal:

begin mathsize 14px style g apostrophe open parentheses x close parentheses equals negative 5 open parentheses 8 x cubed minus x squared plus 3 x minus 5 close parentheses to the power of negative 6 end exponent open parentheses 24 x squared minus 2 x plus 3 close parentheses space space space space space space space space equals open parentheses negative 120 x squared plus 10 x minus 15 close parentheses open parentheses 8 x cubed minus x squared plus 3 x minus 5 close parentheses to the power of negative 6 end exponent space space space space space space space space equals fraction numerator negative 120 x squared plus 10 x minus 15 over denominator open parentheses 8 x cubed minus x squared plus 3 x minus 5 close parentheses to the power of 6 space space space space space space end fraction end style 

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Pembahasan Soal:

Aturan rantai pada turunan:

y equals open square brackets f open parentheses x close parentheses close square brackets to the power of n space rightwards arrow space y apostrophe equals n open square brackets f open parentheses x close parentheses close square brackets to the power of n minus 1 end exponent times f apostrophe open parentheses x close parentheses

Turunan fungsi tersebut dapat ditentukan dengan menggunakan aturan rantai sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row y equals cell square root of 4 x squared minus 5 end root end cell row y equals cell open parentheses 4 x squared minus 5 close parentheses to the power of 1 half end exponent end cell row cell y apostrophe end cell equals cell 1 half open parentheses 4 x squared minus 5 close parentheses to the power of negative 1 half end exponent times open parentheses 8 x close parentheses end cell row cell y apostrophe end cell equals cell 4 x times open parentheses 4 x squared minus 5 close parentheses to the power of negative 1 half end exponent end cell row cell y apostrophe end cell equals cell fraction numerator 4 x over denominator square root of 4 x squared minus 5 end root end fraction end cell end table

Dengan demikian, y apostrophe equals fraction numerator 4 x over denominator square root of 4 x squared minus 5 end root end fraction 

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Pembahasan Soal:

Ingat konsep turunan fungsi komposisi:

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Sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell f left parenthesis x right parenthesis end cell equals cell left parenthesis negative 7 x plus 1 right parenthesis to the power of 5 end cell row cell f apostrophe open parentheses x close parentheses end cell equals cell 5 times negative 7 left parenthesis negative 7 x plus 1 right parenthesis to the power of 5 minus 1 end exponent space end cell row blank equals cell negative 35 left parenthesis negative 7 x plus 1 right parenthesis to the power of 4 end cell row cell f apostrophe apostrophe open parentheses x close parentheses end cell equals cell negative 35 times 4 times negative 7 left parenthesis negative 7 x plus 1 right parenthesis to the power of 4 minus 1 end exponent end cell row blank equals cell 980 left parenthesis negative 7 x plus 1 right parenthesis cubed end cell end table


Jadi, turunan kedua dari fungsi f left parenthesis x right parenthesis equals left parenthesis negative 7 x plus 1 right parenthesis to the power of 5 adalah Error converting from MathML to accessible text..

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Diketahui f(x)=(2x−3)4 dan f(x) adalah turunan pertama dari fungsi f. Nilai f(3) adalah ....

Pembahasan Soal:

Diketahui begin mathsize 14px style f left parenthesis x right parenthesis equals left parenthesis 2 x minus 3 right parenthesis to the power of 4 end style. Fungsi tersebut dapat diturunkan dengan aturan rantai. Ingat bahwa, untuk begin mathsize 14px style y equals open square brackets f left parenthesis x right parenthesis close square brackets to the power of straight n rightwards arrow y apostrophe equals straight n open square brackets straight f left parenthesis straight x right parenthesis close square brackets to the power of straight n minus 1 end exponent cross times straight f apostrophe left parenthesis straight x right parenthesis end style. Diketahui pada soal bahwa fungsi begin mathsize 14px style y equals left parenthesis 2 x minus 3 right parenthesis to the power of 4 end style maka begin mathsize 14px style f apostrophe left parenthesis x right parenthesis equals 2 end style. Turunan pertama begin mathsize 14px style y equals left parenthesis 2 x minus 3 right parenthesis to the power of 4 end style adalah

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell left square bracket f left parenthesis x right parenthesis right square bracket to the power of straight n space end cell row y equals cell left parenthesis 2 x minus 3 right parenthesis to the power of 4 space space end cell row cell y apostrophe end cell equals cell straight n left square bracket straight f left parenthesis straight x right parenthesis right square bracket to the power of straight n minus 1 end exponent cross times straight f apostrophe left parenthesis straight x right parenthesis end cell row blank equals cell 4 left parenthesis 2 straight x minus 3 right parenthesis to the power of 4 minus 1 end exponent cross times 2 space end cell row blank equals cell 4 left parenthesis 2 straight x minus 3 right parenthesis cubed cross times 2 space end cell row blank equals cell 8 left parenthesis 2 straight x minus 3 right parenthesis cubed end cell end table end style

Jadi begin mathsize 14px style f apostrophe left parenthesis x right parenthesis equals 8 left parenthesis 2 x minus 3 right parenthesis cubed end style, maka begin mathsize 14px style f apostrophe left parenthesis 3 right parenthesis end style adalah 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell f apostrophe left parenthesis x right parenthesis end cell equals cell 8 left parenthesis 2 x minus 3 right parenthesis cubed space space end cell row cell f apostrophe left parenthesis 3 right parenthesis space end cell equals cell 8 left parenthesis 2 cross times 3 minus 3 right parenthesis cubed end cell row blank equals cell 8 left parenthesis 6 minus 3 right parenthesis cubed end cell row blank equals cell 8 cross times 3 cubed end cell row blank equals cell 8 cross times 27 end cell row blank equals 216 end table end style

Jadi, jawaban yang tepat adalah E.

1

Roboguru

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