Diketahui : [tex]\bf{y=\cot\left(5x+10\right)}[/tex] Ditanya : [tex]\bf{y'=...}[/tex] [tex]\small\bo.... Question from @Sinogen

JavierSKho13

Penjelasan dengan langkah-langkah:

y = cot (5x + 10)

misal ∅ = 5x + 10

y = cot ∅ → y' = – cosec²∅

∅ = 5x + 10

∅' = 5

maka turunan dari y = cot (5x + 10) adalah

y' = –5 cosec² (5x + 10)

6 votes Thanks 10

henriyulianto

$\large\text{$\begin{aligned}y'=\bf-5\csc^2\left(5x+10\right)\end{aligned}$}$

Pembahasan

Turunan Fungsi Trigonometri

Diberikan fungsi:

$\large\text{$\begin{aligned}y=\cot\left(5x+10\right)\end{aligned}$}$

Kita akan menentukan $y'$ (turunan pertama).

Bentuk $y=\cot\left(5x+10\right)$ adalah bentuk fungsi komposisi $y=(f\circ g)(x)=f(g(x))$ , di mana $f(x)=\cot(g(x))$ dengan $g(x)=5x+10$ .

Dengan aturan rantai fungsi komposisi, $y'$ dapat ditentukan sebagai berikut.

$\begin{aligned}y'&=\left[\cot\left(5x+10\right)\right]'\\&\quad\rightsquigarrow\left[\ \left[\cot(u)\right]'=-\csc^2(u)\ \right]\\&=-\csc^2\left(5x+10\right)\cdot(5x+10)'\\&=-\csc^2\left(5x+10\right)\cdot5\\\therefore\ y'&=\boxed{\ \bf{-}5\csc^2\left(5x+10\right)\ }\end{aligned}$

Cara lainnya:

$\begin{aligned}y'&=\left[\cot\left(5x+10\right)\right]'\\&=\left[\frac{\cos(5x+10)}{\sin(5x+10)}\right]'\\&\ \left[\ \begin{aligned}&\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)\,g(x)-f(x)\,g'(x)}{\left(g(x)\right)^2}\end{aligned}\right.\end{aligned}$
$\begin{aligned}y'&=\frac{\left[\cos(5x+10)\right]'\sin(5x+10)-\cos(5x+10)\left[\sin(5x+10)\right]'}{\sin^2(5x+10)}\\&=\frac{-5\sin(5x+10)\sin(5x+10)-\cos(5x+10)\cdot5\cos(5x+10)}{\sin^2(5x+10)}\\&=\frac{-5\left[\:\sin^2(5x+10)+\cos^2(5x+10)\:\right]}{\sin^2(5x+10)}\\&=\frac{-5(1)}{\sin^2(5x+10)}\\&=-5\cdot\frac{1}{\sin^2(5x+10)}\\y'&=\boxed{\ \bf{-}5\csc^2\left(5x+10\right)\ }\end{aligned}$

$\blacksquare$

13 votes Thanks 25

y' = –5 cosec² (5x + 10)

Pembahasan

Turunan Fungsi Trigonometri

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