Pertanyaan

2. Jika f ( x ) = a sin 2 x + b cos 2 x , tunjukkan bahwa f ( x ) = 2 f ( x ) ( a − b ) sin 2 x

2. Jika $f (x) = a sin^{2} x + b cos^{2} x$ , tunjukkan bahwa $f (x) = \frac{( a - b ) sin 2 x}{2 f ( x )}$

E. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

Jawaban terverifikasi

Jawaban

terbukti bahwa .

terbukti bahwa $f apostrophe open parentheses x close parentheses equals fraction numerator open parentheses a minus b close parentheses space sin space 2 x over denominator 2 space f open parentheses x close parentheses end fraction$ . space

Pembahasan

Turunan fungsi yaitu , turunan fungsi yaitu , dan untuk maka turunannya yaitu . Untuk menyelesaikan soal di atas, perhatikan penghitungan berikut: Jadi, terbukti bahwa .

Turunan fungsi f open parentheses x close parentheses equals sin space g open parentheses x close parentheses yaitu f apostrophe open parentheses x close parentheses equals cos space g open parentheses x close parentheses space g apostrophe open parentheses x close parentheses , turunan fungsi yaitu, dan untuk f open parentheses x close parentheses equals x to the power of n maka turunannya yaitu . Untuk menyelesaikan soal di atas, perhatikan penghitungan berikut:

$f open parentheses x close parentheses equals square root of a space sin squared space x plus b space cos squared space x end root f open parentheses x close parentheses equals open parentheses a space sin squared space x plus b space cos squared space x close parentheses to the power of bevelled 1 half end exponent f apostrophe open parentheses x close parentheses equals 1 half open parentheses a space sin squared space x plus b space cos squared space x close parentheses to the power of bevelled 1 half minus 1 end exponent space open parentheses 2 a space sin space x space cos space x plus 2 b space cos space x space open parentheses negative sin space x close parentheses close parentheses f apostrophe open parentheses x close parentheses equals 1 half open parentheses a space sin squared space x plus b space cos squared space x close parentheses to the power of negative bevelled 1 half end exponent space open parentheses 2 a space sin space x space cos space x minus 2 b space cos space x space sin space x close parentheses f apostrophe open parentheses x close parentheses equals fraction numerator open parentheses 2 a space sin space x space cos space x minus 2 b space cos space x space sin space x close parentheses over denominator 2 open parentheses open parentheses a space sin squared space x plus b space cos squared space x close parentheses to the power of begin display style bevelled 1 half end style end exponent close parentheses end fraction f apostrophe open parentheses x close parentheses equals fraction numerator open parentheses a space 2 space sin space x space cos space x minus b space 2 space cos space x space sin space x close parentheses over denominator 2 open parentheses open parentheses a space sin squared space x plus b space cos squared space x close parentheses to the power of begin display style bevelled 1 half end style end exponent close parentheses end fraction f apostrophe open parentheses x close parentheses equals fraction numerator open parentheses a space sin space 2 x minus b space sin space 2 x close parentheses over denominator 2 open parentheses square root of a space sin squared space x plus b space cos squared space x end root close parentheses end fraction f apostrophe open parentheses x close parentheses equals fraction numerator open parentheses a minus b close parentheses space sin space 2 x over denominator 2 space f open parentheses x close parentheses end fraction$

Jadi, terbukti bahwa $f apostrophe open parentheses x close parentheses equals fraction numerator open parentheses a minus b close parentheses space sin space 2 x over denominator 2 space f open parentheses x close parentheses end fraction$ . space