Pembuktian Rumus Turunan Fungsi Trigonometri - Dokmat

Pembuktian rumus turunan fungsi trigonometri memerlukan tiga syarat materi yaitu kamu harus paham materi turunan fungsi aljabar, sifat limit fungsi trigonometri, dan materi trigonometri itu sendiri. Apabila belum paham, sebaiknya kamu baca dulu materi tersebut!

Rumus turunan fungsi trigonometri adalah suatu rumus yang digunakan untuk menyelesaikan permasalahan turunan trigonometri dengan tujuan agar lebih mudah untuk diselesaikan.

1. Rumus Turunan Fungsi Trigonometri Dasar

Berikut ini adalah rumus turunan fungsi trigonometri yang dapat dipakai untuk menyelesaikan soal turunan fungsi trigonometri.

$y=\sin x\to y^{\prime }=\cos x$

$y=\cos x\to y^{\prime }=-\sin x$

$y=\tan x\to y^{\prime }={\sec }^{2}x$

$y=\cot x\to y^{\prime }=-{\csc }^{2}x$

$y=\sec x\to y^{\prime }=\sec x.\tan x$

$y=\csc x\to y^{\prime }=-\csc x.\cot x$

2. Tips Menghafal Rumus Turunan Trigonometri

Rumus turunan trigonometri memang cukup banyak, tapi jangan khawatir Pak Anwar akan kasih metode untuk mengingatnya. Perhatikan gambar berikut!

Gambar diatas adalah metode untuk mengingat rumus turunan trigonometri, berikut ini adalah penjelasannya.

1). Tanda hasil turunan berselang-seling positif dan negatif dari atas ke bawah (lihat rumus). Gampangkan mengingatnya? Oke lanjut!

2). Rumus pertama dan kedua tinggal di tukar, sin jadi cos dan cos jadi sin.

3). Lihat gambar dan rumus, untuk rumus ketiga dan keempat turunannya sama-sama menggunakan kuadrat dengan melihat hubungan garis pada gambar.

4). Rumus kelima dan keenam, turunanya tinggal tulis dirinya sendiri dikalikan dengan kawannya (hubungan garis pada gambar).

Ini sebenarnya sangat simpel, kenapa tipsnya terlihat panjang? Yaa karena Pak Anwar ngejelasinnya pakai kata-kata.

3. Pembuktian Rumus Turunan Fungsi Trigonometri

Setelah kamu mengetahui rumus turunan fungsi trigonometri, langkah selanjutnya kamu juga harus tau dari mana sih rumus-rumus itu berasal?

Nah berikut ini adalah pembuktian rumus turunan fungsi trigonometri, simak baik-baik yaa!

1. Pembuktian rumus turunan fungsi trigonometri y=sin(x)

Rumus dasar turunan.

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{f(x+h)-f(x)}{h}\right]}$

Rumus pengurangan sinus.

${\displaystyle \sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)}$

Akan Dibuktikan

$y=\sin x\to y^{\prime }=\cos x$

$f(x)=\sin x$ dan $f(x+h)=\sin (x+h)$.

Ingat rumus dasar turunan diatas, sehingga bentuknya menjadi seperti dibawah ini.

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\sin (x+h)-\sin x}{h}\right]}$

Sekarang kita bikin pemisalan aja biar lebih mudah.

Misalkan $A=x+h$ dan $B=x$

Maka

${\displaystyle \begin{array}{rl}\frac{A+B}{2}& =\frac{(x+h)+x}{2}\\ & =\frac{x+h+x}{2}\\ & =\frac{2x+h}{2}\\ & =x+\frac{h}{2}\end{array}}$

Related: Soal Turunan Fungsi Trigonometri

dan

${\displaystyle \begin{array}{rl}\frac{A-B}{2}& =\frac{(x+h)-x}{2}\\ & =\frac{x+h-x}{2}\\ & =\frac{h}{2}\end{array}}$

Kembali ke rumus sebelumnya.

$f^{\prime }(x)={\displaystyle \underset{h\to 0}{lim}\left[\frac{\sin (x+h)-\sin x}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\sin A-\sin B}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{2\cos (x+\frac{h}{2})\sin \left(\frac{h}{2}\right)}{h}\right]}$

${\displaystyle f^{\prime }(x)=2.\underset{h\to 0}{lim}[\cos (x+\frac{h}{2})].\underset{h\to 0}{lim}\left[\frac{\sin \left(\frac{h}{2}\right)}{h}\right]}$

${\displaystyle f^{\prime }(x)=2.\underset{h\to 0}{lim}[\cos (x+\frac{h}{2})].\underset{h\to 0}{lim}\left[\frac{\sin \left(\frac{1}{2}h\right)}{h}\right]}$

(Merah: sifat limit trigonometri)

${\displaystyle f^{\prime }(x)=2.\cos (x+\frac{0}{2}).\frac{1}{2}}$

${\displaystyle f^{\prime }(x)=2.\cos (x+0).\frac{1}{2}}$

${\displaystyle f^{\prime }(x)=2.\cos x.\frac{1}{2}}$

${\displaystyle f^{\prime }(x)=\cos x}$ (terbukti)

2. Pembuktian rumus turunan fungsi trigonometri y=cos(x)

Kita gunakan sebagian rumus yang sama dengan yang sebelumnya, tapi untuk sekarang kita tambah rumus pengurangan cosinus.

${\displaystyle \cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)}$

Akan Dibuktikan

$y=\cos x\to y^{\prime }=-\sin x$

$f(x)=\cos x$ dan $f(x+h)=\cos (x+h)$.

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{f(x+h)-f(x)}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\cos (x+h)-\cos x}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\cos A-\cos B}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{-2\sin (x+\frac{h}{2})\sin \left(\frac{h}{2}\right)}{h}\right]}$

${\displaystyle f^{\prime }(x)=-2.\underset{h\to 0}{lim}[\sin (x+\frac{h}{2})].\underset{h\to 0}{lim}\left[\frac{\sin \left(\frac{h}{2}\right)}{h}\right]}$

${\displaystyle f^{\prime }(x)=-2.\underset{h\to 0}{lim}[\sin (x+\frac{h}{2})].\underset{h\to 0}{lim}\left[\frac{\sin \left(\frac{1}{2}h\right)}{h}\right]}$

(Merah: sifat limit trigonometri)

${\displaystyle f^{\prime }(x)=-2.\sin (x+\frac{0}{2}).\frac{1}{2}}$

${\displaystyle f^{\prime }(x)=-2.\sin (x+0).\frac{1}{2}}$

${\displaystyle f^{\prime }(x)=2.\sin x.\frac{1}{2}}$

${\displaystyle f^{\prime }(x)=-\sin x}$ (terbukti)

3. Pembuktian rumus turunan fungsi trigonometri y=tan(x)

Ingat

${\displaystyle \tan (A\pm B)=\frac{\tan A\pm \tan B}{1\mp \tan A\tan B}}$

Kita akan gunakan pengurangan tangen, maka

${\displaystyle \tan A-\tan B=\tan (A-B).(1+\tan A\tan B)}$

Ingat, Identitas Trigonometri

$1+{\tan }^{2}x={\sec }^{2}x$

Akan Dibuktikan

$y=\tan x\to y^{\prime }={\sec }^{2}x$

$f(x)=\tan x$

$f(x+h)=\tan (x+h)$

Misalkan $A=(x+h)$ dan $B=x$, maka $A-B=h$.

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{f(x+h)-f(x)}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\tan (x+h)-\tan x}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\tan A-\tan B}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\tan (A-B).(1+\tan A\tan B)}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\tan h.(1+\tan (x+h)\tan x)}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\tan h}{h}\right].\underset{h\to 0}{lim}(1+\tan (x+h)\tan x)}$

(merah: sifat limit trigonometri)

${\displaystyle f^{\prime }(x)=1.(1+\tan (x+0)\tan x)}$

${\displaystyle f^{\prime }(x)=1+\tan x.\tan x}$

${\displaystyle f^{\prime }(x)=1+{\tan }^{2}x}$ (identitas trigonometri)

${\displaystyle f^{\prime }(x)={\sec }^{2}x}$ (terbukti)

4. Pembuktian rumus turunan fungsi trigonometri y=cot(x)

Ingat

${\displaystyle \cot \theta =\frac{\cos \theta }{\sin \theta }}$

Ingat

$\sin (A-B)=\sin A\cos B-\cos A\sin B$

Ingat

${\displaystyle \frac{1}{\sin \theta }=\csc \theta }$ berlaku juga untuk ${\displaystyle \frac{1}{{\sin }^2\theta }={\csc }^2\theta }$

Akan Dibuktikan

$y=\cot x\to y^{\prime }=-{\csc }^{2}x$

$f(x)=\cot x$

Related: Soal Turunan Fungsi Trigonometri

$f(x+h)=\cot (x+h)$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{f(x+h)-f(x)}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\cot (x+h)-\cot x}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\frac{\cos (x+h)}{\sin (x+h)}-\frac{\cos x}{\sin x}}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\frac{\sin x\cos (x+h)-\cos x\sin (x+h)}{\sin x\sin (x+h)}}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\sin x\cos (x+h)-\cos x\sin (x+h)}{h\sin x\sin (x+h)}\right]}$

Misalkan $A=x$ dan $B=x+h$

Agar kalian paham, Pak Anwar akan ubah dulu kedalam bentuk A dan B. Tapi kalau udah paham, kalian bisa langsung selesaikan.

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\sin A\cos B-\cos A\sin B}{h\sin A\sin B}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\sin (A-B)}{h\sin A\sin B}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\sin (x-(x+h))}{h\sin x\sin (x+h)}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\sin (-h)}{h\sin x\sin (x+h)}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\sin (-h)}{h}\right].\underset{h\to 0}{lim}\left[\frac{1}{\sin x}\right].\underset{h\to 0}{lim}\left[\frac{1}{\sin (x+h)}\right]}$

${\displaystyle f^{\prime }(x)=-1.\frac{1}{\sin x}.\frac{1}{\sin (x+0)}}$

${\displaystyle f^{\prime }(x)=-\frac{1}{\sin x}.\frac{1}{\sin x}}$

${\displaystyle f^{\prime }(x)=-\frac{1}{{\sin }^{2}x}}$

${\displaystyle f^{\prime }(x)=-{\csc }^{2}x}$ (terbukti)

5. Pembuktian rumus turunan fungsi trigonometri y=sec(x)

Ingat

${\displaystyle \cos \theta =\frac{1}{\sec \theta }\to \sec \theta =\frac{1}{\cos \theta }}$

Ingat

${\displaystyle \tan \theta =\frac{\sin \theta }{\cos \theta }}$

Ingat

${\displaystyle \cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)}$

Akan Dibuktikan

$y=\sec x\to y^{\prime }=\sec x.\tan x$

$f(x)=\sec x$

$f(x+h)=\sec (x+h)$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{f(x+h)-f(x)}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\sec (x+h)-\sec x}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\frac{1}{\cos (x+h)}-\frac{1}{\cos x}}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\frac{\cos x-\cos (x+h)}{\cos (x+h)\cos x}}{h}\right]}$

Misalkan $A=x$ dan $B=x+h$, maka ${\displaystyle \frac{A+B}{2}=x+\frac{h}{2}}$ dan ${\displaystyle \frac{A-B}{2}=-\frac{h}{2}}$

Sehingga,

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\frac{\cos A-\cos B}{\cos B\cos A}}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\frac{-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)}{\cos B\cos A}}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\frac{-2\sin (x+\frac{h}{2})\sin (-\frac{h}{2})}{\cos (x+h)\cos x}}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{-2\sin (x+\frac{h}{2})\sin (-\frac{h}{2})}{\cos (x+h).\cos x.h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\sin (x+\frac{h}{2})}{\cos (x+h)}\right].\underset{h\to 0}{lim}\left[\frac{\sin (-\frac{1}{2}h)}{h}\right].\underset{h\to 0}{lim}\left[\frac{-2}{\cos x}\right]}$

${\displaystyle f^{\prime }(x)=\frac{\sin (x+\frac{0}{2})}{\cos (x+0)}.(-\frac{1}{2}).\left(\frac{-2}{\cos x}\right)}$

${\displaystyle f^{\prime }(x)=\frac{\sin x}{\cos x}.\frac{1}{\cos x}}$

${\displaystyle f^{\prime }(x)=\tan x.\sec x}$

${\displaystyle f^{\prime }(x)=\sec x.\cos x}$ (terbukti)

6. Pembuktian rumus turunan fungsi trigonometri y=csc(x)

Ingat

${\displaystyle \sin \theta =\frac{1}{\csc \theta }\to \csc \theta =\frac{1}{\sin \theta }}$

Ingat

${\displaystyle \cot \theta =\frac{\cos \theta }{\sin \theta }}$

Ingat

${\displaystyle \sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)}$

Akan Dibuktikan

$y=\csc x\to y^{\prime }=-\csc x.\cot x$

$f(x)=\csc x$

$f(x+h)=\csc (x+h)$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{f(x+h)-f(x)}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\csc (x+h)-\csc x}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\frac{1}{\sin (x+h)}-\frac{1}{\sin x}}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\frac{\sin x-\sin (x+h)}{\sin (x+h)\sin x}}{h}\right]}$

Misalkan $A=x$ dan $B=x+h$, maka ${\displaystyle \frac{A+B}{2}=x+\frac{h}{2}}$ dan ${\displaystyle \frac{A-B}{2}=-\frac{h}{2}}$

Sehingga,

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\frac{\sin A-\sin B}{\sin B\sin A}}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\frac{2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)}{\sin B\sin A}}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\frac{2\cos (x+\frac{h}{2})\sin (-\frac{h}{2})}{\sin (x+h)\sin x}}{h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{2\cos (x+\frac{h}{2})\sin (-\frac{h}{2})}{\sin (x+h).\sin x.h}\right]}$

${\displaystyle f^{\prime }(x)=\underset{h\to 0}{lim}\left[\frac{\cos (x+\frac{h}{2})}{\sin (x+h)}\right].\underset{h\to 0}{lim}\left[\frac{\sin (-\frac{1}{2}h)}{h}\right].\underset{h\to 0}{lim}\left[\frac{2}{\sin x}\right]}$

${\displaystyle f^{\prime }(x)=\frac{\cos (x+\frac{0}{2})}{\sin (x+0)}.(-\frac{1}{2}).\left(\frac{2}{\sin x}\right)}$

${\displaystyle f^{\prime }(x)=-\frac{\cos x}{\sin x}.\frac{1}{\sin x}}$

${\displaystyle f^{\prime }(x)=-\cot x.\csc x}$

${\displaystyle f^{\prime }(x)=-\csc x.\cot x}$ (terbukti)

Itulah pembahasan lengkap pembuktian rumus turunan fungsi trigonometri, semoga artikel rumus turunan fungsi trigonometri ini bermanfaat untuk banyak orang. Oleh karena itu, tolong bagikan sebanyak-banyaknya.