Selanjutnya saat kita masih kukuh menggu nakan rumus semual, maka saat menentukan turunan pertama fungsi $\tan x$, kita akan ketemu bentuk $\color{blue}\sin (x+h)\cos x$ dan $\color{blue}\cos (x+h)\sin x$, maka saat ketemu bentuk itu kita gunakan rumus:
$\color{purple}\begin{cases} \sin A\cos B&=\displaystyle \frac{1}{2}\left ( \sin (A+B)+\sin (A-B) \right ) \\\\ \cos A\sin B&=\displaystyle \frac{1}{2}\left ( \sin (A+B)-\sin (A-B) \right ) \end{cases}$
Coba perhatikanlah uraian turunan fungsi tangen berikut:
$\color{purple}\begin{aligned}f'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan (x+h)-\tan x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{ \sin (x+h)}{\cos (x+h)}-\displaystyle \frac{\sin x}{\cos x}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{ \sin (x+h)\cos x}{\cos (x+h)\cos x}-\displaystyle \frac{\cos (x+h)\sin x}{\cos (x+h)\cos x}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{\cos (x+h).\cos x.h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{...+\displaystyle \frac{1}{2}\sin h-...+\displaystyle \frac{1}{2}\sin h}{\cos (x+h).\cos x.h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin h}{h.\cos (x+h).\cos x}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \left ( \displaystyle \frac{\sin h}{h} \right )\left ( \displaystyle \frac{1}{\cos (x+h)\cos x} \right )\\ &=1\times \displaystyle \frac{1}{\cos (x+0).\cos x}\\ &=\displaystyle \frac{1}{\cos ^{2}x}\\ &=\sec ^{2}x \end{aligned}$
Atau kita juga dapat menggunakan rumus $\color{red}\sin (A-B)=\sin A\cos B-\cos A\sin B$ sebagaimana berikut ini (perhatikanlah proses langkah 5 ke langkah 6):
$\color{blue}\begin{aligned}f'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan (x+h)-\tan x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{ \sin (x+h)}{\cos (x+h)}-\displaystyle \frac{\sin x}{\cos x}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{ \sin (x+h)\cos x}{\cos (x+h)\cos x}-\displaystyle \frac{\cos (x+h)\sin x}{\cos (x+h)\cos x}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{\cos (x+h).\cos x.h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin \left ((x+h)-x \right )}{\cos (x+h).\cos x.h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin h}{h.\cos (x+h).\cos x}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \left ( \displaystyle \frac{\sin h}{h} \right )\left ( \displaystyle \frac{1}{\cos (x+h)\cos x} \right )\\ &=1\times \displaystyle \frac{1}{\cos (x+0).\cos x}\\ &=\displaystyle \frac{1}{\cos ^{2}x}\\ &=\sec ^{2}x \end{aligned}$
Berikut hasil turunan pertama untuk fungsi trigonometri yang perlu diingat:
$\color{purple}\begin{aligned}1.\quad &f(x)=\sin x\Rightarrow f'(x)=\cos x\\ 2.\quad &f(x)=\cos x\Rightarrow f'(x)=-\sin x\\ 3.\quad &f(x)=\tan x\Rightarrow f'(x)=\sec ^{2}x\\ 4.\quad &f(x)=\cot x\Rightarrow f'(x)=-\csc ^{2}x\\ 5.\quad &f(x)=\sec x\Rightarrow f'(x)=\sec x\tan x\\ 6.\quad &f(x)=\csc x\Rightarrow f'(x)=-\csc x\cot x \end{aligned}$
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