Pertanyaan

Jika maka

Jika $f open parentheses x close parentheses equals fraction numerator 1 plus tan invisible function application x over denominator tan invisible function application x end fraction$ maka

-2
1
2

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I. Roy

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Pembahasan

$f open parentheses x close parentheses equals u over v u equals 1 plus tan space x u to the power of apostrophe equals sec squared x v equals tan space x v to the power of apostrophe equals sec squared x f to the power of apostrophe open parentheses x close parentheses equals fraction numerator u to the power of apostrophe v minus u space v apostrophe over denominator v squared end fraction space space space space space space space equals fraction numerator sec squared x open parentheses tan space x close parentheses minus left parenthesis 1 plus tan space x right parenthesis sec squared x over denominator tan squared x end fraction space space space space space space space equals negative fraction numerator sec squared x over denominator tan squared x end fraction f to the power of apostrophe open parentheses pi over 4 close parentheses equals negative fraction numerator sec squared pi over 4 over denominator tan squared pi over 4 end fraction space space space space space space space space space space space equals negative open parentheses square root of 2 close parentheses squared over 1 squared equals negative 2$