Pertanyaan

Turunan petama dari fungsi g ( t ) = 1 + cos 2 t 1 − sin 2 t adalah ....

Turunan petama dari fungsi $g (t) = \frac{1 - sin 2 t}{1 + cos 2 t}$ adalah ....

$begin mathsize 14px style g apostrophe open parentheses t close parentheses equals fraction numerator 2 sin invisible function application 2 t minus 2 cos invisible function application 2 t over denominator left parenthesis 1 plus c o s invisible function application 2 t right parenthesis squared end fraction end style$
$begin mathsize 14px style g apostrophe open parentheses t close parentheses equals fraction numerator 2 cos invisible function application 2 t minus 2 sin invisible function application 2 t minus 2 over denominator left parenthesis 1 plus c o s invisible function application 2 t right parenthesis squared end fraction end style$
$begin mathsize 14px style g apostrophe open parentheses t close parentheses equals fraction numerator 2 sin invisible function application 2 t minus 2 cos invisible function application 2 t minus 2 over denominator left parenthesis 1 plus c o s invisible function application 2 t right parenthesis squared end fraction end style$

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Klaim

H. Nufus

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

Pembahasan

Misal dan . Maka dan . Ingat kembali jika , maka . Sehingga,didapat Jadi, jawaban yang tepat adalah E.

Misal dan . Maka dan .

Ingat kembali jika $begin mathsize 14px style g left parenthesis t right parenthesis equals fraction numerator u left parenthesis t right parenthesis over denominator v left parenthesis t right parenthesis space end fraction end style$ , maka $begin mathsize 14px style g to the power of apostrophe left parenthesis t right parenthesis equals fraction numerator u to the power of apostrophe left parenthesis t right parenthesis times v left parenthesis t right parenthesis minus u left parenthesis t right parenthesis times v to the power of apostrophe left parenthesis t right parenthesis over denominator left parenthesis v left parenthesis t right parenthesis right parenthesis squared end fraction end style$ .

Sehingga, didapat

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell g to the power of apostrophe left parenthesis t right parenthesis end cell equals cell fraction numerator u to the power of apostrophe left parenthesis t right parenthesis times v left parenthesis t right parenthesis minus u left parenthesis t right parenthesis times v to the power of apostrophe left parenthesis t right parenthesis over denominator left parenthesis v left parenthesis t right parenthesis right parenthesis squared end fraction end cell row blank equals cell fraction numerator negative 2 cos invisible function application 2 t times left parenthesis 1 plus cos invisible function application 2 t right parenthesis minus left parenthesis 1 minus sin invisible function application 2 t right parenthesis times left parenthesis negative 2 sin invisible function application 2 t right parenthesis over denominator left parenthesis 1 plus c o s invisible function application 2 t right parenthesis squared end fraction end cell row blank equals cell fraction numerator negative 2 cos invisible function application 2 t minus 2 cos squared invisible function application 2 t minus left parenthesis negative 2 sin invisible function application 2 t plus 2 sin squared invisible function application 2 t right parenthesis over denominator left parenthesis 1 plus c o s invisible function application 2 t right parenthesis squared end fraction end cell row blank equals cell fraction numerator negative 2 cos invisible function application 2 t minus 2 cos squared invisible function application 2 t plus 2 sin invisible function application 2 t minus 2 sin squared invisible function application 2 t over denominator left parenthesis 1 plus c o s invisible function application 2 t right parenthesis squared end fraction end cell row blank equals cell fraction numerator 2 sin invisible function application 2 t minus 2 cos invisible function application 2 t minus 2 left parenthesis sin squared invisible function application 2 t plus cos squared invisible function application 2 t right parenthesis over denominator left parenthesis 1 plus c o s invisible function application 2 t right parenthesis squared end fraction end cell row blank equals cell fraction numerator 2 sin invisible function application 2 t minus 2 cos invisible function application 2 t minus 2 over denominator left parenthesis 1 plus c o s invisible function application 2 t right parenthesis squared end fraction end cell end table end style$

Jadi, jawaban yang tepat adalah E.