Pertanyaan

Turunan pertama dari fungsi f ( x ) = ln x ⋅ lo g x adalah ....

Turunan pertama dari fungsi $f (x) = ln x \cdot lo g x$ adalah ....

$begin mathsize 12px style f apostrophe open parentheses x close parentheses equals fraction numerator ln space x over denominator x end fraction end style$
$begin mathsize 12px style f apostrophe open parentheses x close parentheses equals fraction numerator 2 ln space x over denominator x end fraction end style$
$begin mathsize 12px style f apostrophe open parentheses x close parentheses equals fraction numerator 2 over denominator x ln space x end fraction end style$
$begin mathsize 12px style f apostrophe open parentheses x close parentheses equals fraction numerator log space x over denominator x end fraction end style$
$begin mathsize 12px style f apostrophe open parentheses x close parentheses equals fraction numerator 2 log space x over denominator x end fraction end style$

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

Pembahasan

Misalkan u(x) = ln ⁡x dan . Didapat bahwa dan . Oleh karena itu, didapat Ingat bahwa . Oleh karena itu, Jadi, jawaban yang tepat adalah E.

Misalkan u(x) = ln ⁡x dan . Didapat bahwa dan $begin mathsize 12px style v to the power of apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator x ln invisible function application 10 end fraction end style$ .

Oleh karena itu, didapat

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell ln invisible function application x times log invisible function application x end cell row cell f open parentheses x close parentheses end cell equals cell u open parentheses x close parentheses v open parentheses x close parentheses end cell row cell f to the power of apostrophe open parentheses x close parentheses end cell equals cell u to the power of apostrophe open parentheses x close parentheses v open parentheses x close parentheses plus u open parentheses x close parentheses v to the power of apostrophe open parentheses x close parentheses end cell row cell f to the power of apostrophe open parentheses x close parentheses end cell equals cell 1 over x times log invisible function application x plus ln invisible function application x times fraction numerator 1 over denominator x ln invisible function application 10 end fraction end cell end table end style$

Ingat bahwa $begin mathsize 12px style fraction numerator ln invisible function application a over denominator ln invisible function application b end fraction equals log presubscript blank presuperscript b invisible function application a end style$ . Oleh karena itu,

$begin mathsize 12px style f to the power of apostrophe open parentheses x close parentheses equals 1 over x times log invisible function application x plus ln invisible function application x times fraction numerator 1 over denominator x ln invisible function application 10 end fraction f to the power of apostrophe open parentheses x close parentheses equals 1 over x times log invisible function application x plus 1 over x times fraction numerator ln invisible function application x over denominator ln invisible function application 10 end fraction f to the power of apostrophe open parentheses x close parentheses equals 1 over x times log invisible function application x plus 1 over x times log presubscript blank presuperscript 10 invisible function application x f to the power of apostrophe open parentheses x close parentheses equals 1 over x times log invisible function application x plus 1 over x times log invisible function application x f to the power of apostrophe open parentheses x close parentheses equals 2 times 1 over x times log invisible function application x f to the power of apostrophe open parentheses x close parentheses equals fraction numerator 2 log invisible function application x over denominator x end fraction end style$

Jadi, jawaban yang tepat adalah E.