Lanjutan Materi (3) Turunan fungsi Trigonometri (Matematika Peminatan Kelas XII)

Selanjutnya saat kita masih kukuh menggu nakan rumus semual, maka saat menentukan turunan pertama fungsi  $\tan x$, kita akan ketemu bentuk $\sin (x+h)\cos x$  dan $\cos (x+h)\sin x$, maka saat ketemu bentuk itu kita gunakan rumus:

$\{\begin{array}{ll}\sin A\cos B& ={\displaystyle \frac{1}{2}(\sin (A+B)+\sin (A-B))}\\ \\ \cos A\sin B& ={\displaystyle \frac{1}{2}(\sin (A+B)-\sin (A-B))}\end{array}$

Coba perhatikanlah uraian turunan fungsi tangen berikut:

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\tan (x+h)-\tan x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{\sin (x+h)}{\cos (x+h)}-{\displaystyle \frac{\sin x}{\cos x}}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{\sin (x+h)\cos x}{\cos (x+h)\cos x}-{\displaystyle \frac{\cos (x+h)\sin x}{\cos (x+h)\cos x}}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{\cos (x+h).\cos x.h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{...+{\displaystyle \frac{1}{2}\sin h-...+{\displaystyle \frac{1}{2}\sin h}}}{\cos (x+h).\cos x.h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin h}{h.\cos (x+h).\cos x}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \left({\displaystyle \frac{\sin h}{h}}\right)\left({\displaystyle \frac{1}{\cos (x+h)\cos x}}\right)}\\ & =1\times {\displaystyle \frac{1}{\cos (x+0).\cos x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}}\\ & ={\sec }^{2}x\end{array}$

Atau kita juga dapat menggunakan rumus $\sin (A-B)=\sin A\cos B-\cos A\sin B$ sebagaimana berikut ini (perhatikanlah proses langkah 5 ke langkah 6):

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\tan (x+h)-\tan x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{\sin (x+h)}{\cos (x+h)}-{\displaystyle \frac{\sin x}{\cos x}}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{\sin (x+h)\cos x}{\cos (x+h)\cos x}-{\displaystyle \frac{\cos (x+h)\sin x}{\cos (x+h)\cos x}}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{\cos (x+h).\cos x.h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin ((x+h)-x)}{\cos (x+h).\cos x.h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin h}{h.\cos (x+h).\cos x}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \left({\displaystyle \frac{\sin h}{h}}\right)\left({\displaystyle \frac{1}{\cos (x+h)\cos x}}\right)}\\ & =1\times {\displaystyle \frac{1}{\cos (x+0).\cos x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}}\\ & ={\sec }^{2}x\end{array}$

Berikut hasil turunan pertama untuk fungsi trigonometri yang perlu diingat:

$\begin{array}{rl}1.& f(x)=\sin x\Rightarrow f^{\prime }(x)=\cos x\\ 2.& f(x)=\cos x\Rightarrow f^{\prime }(x)=-\sin x\\ 3.& f(x)=\tan x\Rightarrow f^{\prime }(x)={\sec }^{2}x\\ 4.& f(x)=\cot x\Rightarrow f^{\prime }(x)=-{\csc }^{2}x\\ 5.& f(x)=\sec x\Rightarrow f^{\prime }(x)=\sec x\tan x\\ 6.& f(x)=\csc x\Rightarrow f^{\prime }(x)=-\csc x\cot x\end{array}$