Turunan Fungsi Trigonometri

Turunan Fungsi Sinus

Misalkan diketahui fungsi sinus $f\left( x \right) = \sin x$. Turunan fungsi sinus $f\left( x \right) = \sin x$ ditentukan sebagai berikut:

$\begin{array}{l} f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {x + h} \right) - \sin x}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( x \right)\cos \left( h \right) + \cos \left( x \right)\sin \left( h \right) - \sin x}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\sin x\left[ {\cos \left( h \right) - 1} \right] + \cos \left( x \right)\sin \left( h \right)}}{h}\\ &= \sin x\mathop {\lim }\limits_{h \to 0} \frac{{\cos \left( h \right) - 1}}{h} + \cos x\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( h \right)}}{h} \end{array}$

Berdasarkan perhitungan limit fungsi trigonometri dapat ditunjukkan bahwa $\mathop {\lim }\limits_{h \to 0} \frac{{\cos \left( h \right) - 1}}{h} = 0$ dan $\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( h \right)}}{h} = 1$. Dengan substitusi nilai-nilai tersebut ke $f'\left( x \right)$ diperoleh

$\begin{array}{l} f'\left( x \right) &= \sin x\mathop {\lim }\limits_{h \to 0} \frac{{\cos \left( h \right) - 1}}{h} + \cos x\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( h \right)}}{h}\\ &= \sin x.0 + \cos x.1\\ &= \cos x \end{array}$

Jadi, dapat disimpulkan bahwa:

Jika $f\left( x \right) = \sin x$ maka $f'\left( x \right) = \cos x$.

Turunan Fungsi Cosinus

Misalkan diketahui fungsi cosinus $f\left( x \right) = \cos x$. Turunan fungsi cosinus $f\left( x \right) = \cos x$ ditentukan sebagai berikut:

$\begin{array}{l} f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\cos \left( {x + h} \right) - \cos x}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\cos \left( x \right)\cos \left( h \right) - \sin \left( x \right)\sin \left( h \right) - \cos x}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\cos x\left[ {\cos \left( h \right) - 1} \right] - \sin \left( x \right)\sin \left( h \right)}}{h}\\ &= \cos x\mathop {\lim }\limits_{h \to 0} \frac{{\cos \left( h \right) - 1}}{h} - \sin x\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( h \right)}}{h}\\ &= \cos x.0 - \sin x.1\\ &= - \sin x \end{array}$

Jadi, dapat disimpulkan bahwa:

Jika $f\left( x \right) = \cos x$ maka $f'\left( x \right) = - \sin x$.

Turunan Fungsi Tangen

Misalkan diketahui fungsi $f\left( x \right) = \tan x$. Oleh karena $\tan x = \frac{{\sin x}}{{\cos x}}$, maka fungsi $f\left( x \right) = \tan x = \frac{{\sin x}}{{\cos x}}$ dengan $\cos x \ne 0$ merupakan hasil bagi $u\left( x \right) = \sin x$ dengan $v\left( x \right) = \cos x$.

• $u\left( x \right) = \sin x$ maka $u'\left( x \right) = \cos x$.
• $v\left( x \right) = \cos x$ maka $v'\left( x \right) = - \sin x$.

Dengan menggunakan rumus turunan fungsi sinus dan rumus turunan fungsi cosinus, turunan fungsi $f\left( x \right) = \tan x = \frac{{\sin x}}{{\cos x}}$ dapat ditentukan dengan rumus hasil bagi fungsi-fungsi sebagai berikut:

$\begin{array}{l} f'\left( x \right) &= \frac{{u'\left( x \right).v\left( x \right) - v'\left( x \right).u\left( x \right)}}{{{{\left[ {v\left( x \right)} \right]}^2}}}\\ &= \frac{{\cos x.\cos x - \left( { - \sin x} \right).\sin x}}{{{{\left[ {\cos x} \right]}^2}}}\\ &= \frac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}}\\ &= \frac{1}{{{{\cos }^2}x}}\\ &= {\sec ^2}x \end{array}$

Jadi, dapat disimpulkan bahwa

Jika $f\left( x \right) = \tan x$ maka $f'\left( x \right) = {\sec ^2}x$.

Turunan Fungsi Cotangen, Secan, dan Cosecan

Turunan fungsi Cotangen, Fungsi Secan, dan Fungsi Cosecan masing-masing disajikan sebagai berikut:

1. Jika $f\left( x \right) = \cot x$ maka $f'\left( x \right) = - {\text{cosec}^2}x$
2. Jika $f\left( x \right) = \sec x$ maka $f'\left( x \right) = \sec x.\tan x$
3. Jika $f\left( x \right) = \text{cosec}x$ maka $f'\left( x \right) = - \text{cosec}x.\cot x$

Latihan-Latihan

Contoh

Carilah turunan atau $f'\left( x \right)$ untuk fungsi-fungsi trigonometri berikut:

1. $f\left( x \right) = 2\sin x + 3\cos x$
2. $f\left( x \right) = 4\sin x\cos x$
3. $f\left( x \right) = \frac{{\sin x}}{{\sin x - \cos x}}$
4. $f\left( x \right) = \sqrt {1 + \sin x} $
5. $f\left( x \right) = \frac{{2\cos x}}{{{x^2}}}$

Jawab

1. $f\left( x \right) = 2\sin x + 3\cos x$
$f'\left( x \right) = 2\cos x - 3\sin x$
2. $f\left( x \right) = 4\sin x\cos x$
$\begin{array}{l} f\left( x \right) &= 2\left( {2\sin x\cos x} \right)\\ &= 2\sin 2x\\ f'\left( x \right) &= 2.2.\cos 2x\\ &= 4\cos 2x \end{array}$
3. $f\left( x \right) = \frac{{\sin x}}{{\sin x - \cos x}}$
$\begin{array}{l} u\left( x \right) &= \sin x \Rightarrow u'\left( x \right) = \cos x\\ v\left( x \right) &= \sin x - \cos x \Rightarrow v'\left( x \right) = \cos x + \sin x\\ f\left( x \right) &= \frac{{u'\left( x \right).v\left( x \right) - v'\left( x \right).u\left( x \right)}}{{{{\left[ {v\left( x \right)} \right]}^2}}}\\ &= \frac{{\cos x.\left( {\sin x - \cos x} \right) - \left( {\cos x + \sin x} \right).\sin x}}{{{{\left( {\sin x - \cos x} \right)}^2}}}\\ &= \frac{{\cos x\sin x - {{\cos }^2}x - \cos x\sin x - {{\sin }^2}x}}{{{{\left( {\sin x - \cos x} \right)}^2}}}\\ &= \frac{{ - \left( {{{\cos }^2}x + {{\sin }^2}x} \right)}}{{{{\left( {\sin x - \cos x} \right)}^2}}}\\ &= \frac{{ - 1}}{{\left( {{{\sin }^2}x + {{\cos }^2}x - 2\sin x\cos x} \right)}}\\ &= - \frac{1}{{1 - \sin 2x}} \end{array}$
4. $f\left( x \right) = \sqrt {1 + \sin x} $
$\begin{array}{l} f\left( x \right) &= {\left( {1 + \sin x} \right)^{\frac{1}{2}}}\\ f'\left( x \right) &= \frac{1}{2}.\cos x.{\left( {1 + \sin x} \right)^{\frac{1}{2} - 1}}\\ &= \frac{1}{2}.\cos x.{\left( {1 + \sin x} \right)^{ - \frac{1}{2}}}\\ &= \frac{{\cos x}}{{2{{\left( {1 + \sin x} \right)}^{\frac{1}{2}}}}}\\ &= \frac{{\cos x}}{{2\sqrt {1 + \sin x} }} \end{array}$
5. $f\left( x \right) = \frac{{2\cos x}}{{{x^2}}}$
$\begin{array}{l} u\left( x \right) &= 2\cos x \Rightarrow u'\left( x \right) = - 2\sin x\\ v\left( x \right) &= {x^2} \Rightarrow v'\left( x \right) = 2x\\ f\left( x \right) &= \frac{{u'\left( x \right).v\left( x \right) - v'\left( x \right).u\left( x \right)}}{{{{\left[ {v\left( x \right)} \right]}^2}}}\\ &= \frac{{ - 2\sin x.{x^2} - 2x.2\cos x}}{{{{\left( {{x^2}} \right)}^2}}}\\ &= \frac{{ - 2{x^2}\sin x - 4x\cos x}}{{{x^4}}}\\ &= \frac{{ - 2x\sin x - 4\cos x}}{{{x^3}}} \end{array}$