Pertanyaan

Diketahui , maka nilai ....

S. Nur

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

Pembahasan

Misalkan , maka . Sehingga turunan pertama dari fungsi tersebut yaitu Maka nilai dari adalah Jadi, jawaban yang tepat adalah C.

Misalkan , maka .

Sehingga turunan pertama dari fungsi tersebut yaitu

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight f open parentheses straight x close parentheses end cell equals cell straight u to the power of 1 third end exponent straight f to the power of straight apostrophe open parentheses straight x close parentheses equals 1 third straight u to the power of negative 2 over 3 end exponent times straight u straight apostrophe end cell row cell straight f to the power of straight apostrophe open parentheses straight x close parentheses end cell equals cell open parentheses 1 third open parentheses straight x to the power of 4 plus 5 straight x squared minus 7 close parentheses to the power of negative 2 over 3 end exponent close parentheses open parentheses 4 straight x cubed plus 10 straight x close parentheses end cell row cell straight f to the power of straight apostrophe open parentheses straight x close parentheses end cell equals cell fraction numerator open parentheses 4 straight x cubed plus 10 straight x close parentheses over denominator 3 cube root of open parentheses straight x to the power of 4 plus 5 straight x squared minus 7 close parentheses squared end root end fraction end cell end table end style$

Maka nilai dari adalah

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight f to the power of straight apostrophe open parentheses 1 close parentheses plus straight f to the power of straight apostrophe open parentheses negative 1 close parentheses end cell equals cell fraction numerator open parentheses 4 open parentheses 1 close parentheses cubed plus 10 open parentheses 1 close parentheses close parentheses over denominator 3 cube root of open parentheses 1 to the power of 4 plus 5 open parentheses 1 close parentheses squared minus 7 close parentheses squared end root end fraction plus fraction numerator open parentheses 4 open parentheses negative 1 close parentheses cubed plus 10 open parentheses negative 1 close parentheses close parentheses over denominator 3 cube root of open parentheses open parentheses negative 1 close parentheses to the power of 4 plus 5 open parentheses negative 1 close parentheses squared minus 7 close parentheses squared end root end fraction end cell row blank equals cell fraction numerator 14 over denominator 3 open parentheses 1 close parentheses end fraction minus fraction numerator 14 over denominator 3 open parentheses 1 close parentheses end fraction end cell row blank equals 0 end table end style$

Jadi, jawaban yang tepat adalah C.