Jika [tex]\displaystyle f(x)=\frac{\sin x-\cos x}{\sin x+\cos x}[/tex] tentukan [tex]\displaystyle f'\left

Berikut ini adalah pertanyaan dari syakhayaz pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Jika $\displaystyle f(x)=\frac{\sin x-\cos x}{\sin x+\cos x}$ tentukan $\displaystyle f'\left ( \frac{\pi}{12} \right )$

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawab:
$\large\boxed{ f'\left(\frac{\pi}{12}\right)=\frac{4}{3}}$

Penjelasan dengan langkah-langkah:
$\displaystyle\tt f(x)=\frac{sin x-cosx}{sinx + cosx}$
$\displaystyle\tt f(x)=\frac{u}{v} ~- > ~f'(x) = \frac{u'v-uv'}{v^2}$
$\tt f'(x) = \frac{(sinx-cosx)'(sinx+cosx)-(sinx-cosx)(sinx+cosx)'}{(sinx+cosx)^2}\\\\f'(x) = \frac{(sinx+cosx)(sinx+cosx)-(sinx-cosx)(cosx-sinx)}{(sinx+cosx)^2}\\\\f'(x) = \frac{(sinx+cosx)(sinx+cosx)+(sinx-cosx)(sinx-cosx)}{(sinx+cosx)^2}$
$\displaystyle\tt f'(x) = \frac{(sinx+cosx)^2+(sinx-cosx)^2}{(sinx+cosx)^2}$
$\tt f'(x) = \frac{cos^2x+sin^2x+2sinxcosx+cos^2x+sin^2x-2sinxcosx}{cos^2x+sin^2x+2sinxcosx}$
$\displaystyle\tt f'(x)=\frac{1+1}{1+sin(2x)}$
$\displaystyle\tt f'(x)=\frac{2}{1+sin(2x)}$
$\displaystyle\tt f'\left(\frac{\pi}{12}\right)=\frac{2}{1+sin\left(2\left(\frac{\pi}{12}\right)\right)}$
$\displaystyle\tt f'\left(\frac{\pi}{12}\right)=\frac{2}{1+sin\left(\frac{\pi}{6}\right)}$
$\displaystyle\tt f'\left(\frac{\pi}{12}\right)=\frac{2}{1+\frac{1}{2}}$
$\displaystyle\tt f'\left(\frac{\pi}{12}\right)=\frac{2}{\frac{2}{2}+\frac{1}{2}}$
$\displaystyle\tt f'\left(\frac{\pi}{12}\right)=\frac{2}{\frac{3}{2}}$
$\displaystyle\tt f'\left(\frac{\pi}{12}\right)=\frac{2(2)}{3}$
$\large\boxed{ f'\left(\frac{\pi}{12}\right)=\frac{4}{3}}$