Exercise Zone: Turunan Fungsi Trigonometri [2]

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Tipe:

Standar SBMPTN HOTS

No.

Diketahui {f(x)=\cos^2x+4\cos x+4} dan {g(x)=\cos x+2}.

1. Tentukan turunan pertama dari h(x), jika diketahui h(x)=\dfrac{f(x)}{g(x)}
2. Hitunglah nilai dari \dfrac{d^2h(x)}{dx^2} untuk x=\dfrac{\pi}3

Grup Telegram

1. h(x)=\dfrac{f(x)}{g(x)}
   \begin{aligned} h(x)&=\dfrac{\cos^2x+4\cos x+4}{\cos x+2}\\ &=\dfrac{(\cos x+2)^2}{\cos x+2}\\ &=\cos x+2\\ \dfrac{dh(x)}{dx}&=\boxed{\boxed{-\sin x}} \end{aligned}
   
   

2. \dfrac{d^2h(x)}{dx^2}=-\cos x
   \begin{aligned} \dfrac{d^2h\left(\dfrac{\pi}3\right)}{dx^2}&=-\cos\dfrac{\pi}3\\ &=\boxed{\boxed{-\dfrac12}} \end{aligned}

Jadi, turunan pertama dari h(x) adalah -\sin x nilai dari \dfrac{d^2h(x)}{dx^2} untuk x=\dfrac{\pi}3 adalah -\dfrac12

No.

Tentukan y' jlka diketahui fungsi y sebagai berikut!
y=x^3\tan2x-\dfrac12x^2\tan2x+x\tan2x-\tan2x

Grup Telegram

\begin{aligned} y&=x^3\tan2x-\dfrac12\tan2x+x\tan2x-\tan2x\\ &=\tan2x\left(x^3-\dfrac12x^2+x-1\right) \end{aligned}

\begin{aligned} u&=\tan2x\\ u'&=2\sec^22x \end{aligned}

\begin{aligned} v&=x^3-\dfrac12x^2+x-1\\ v'&=3x^2-x+1 \end{aligned}

\begin{aligned} y'&=u'v+uv'\\ &=2\sec^22x\left(x^3-\dfrac12x^2+x-1\right)+\tan2x\left(3x^2-x+1\right)\\ &=\boxed{\boxed{2x^3\sec^22x-x^2\sec^22x+2x\sec^22x-2\sec^22x+3x^2\tan2x-x\tan2x+\tan2x}} \end{aligned}

Jadi, y'=2x^3\sec^22x-x^2\sec^22x+2x\sec^22x-2\sec^22x+3x^2\tan2x-x\tan2x+\tan2x.

No.

Tentukan turunan dari fungsi trigonometri berikut!
y=\sin10x

Grup Telegram

\begin{aligned} y&=\sin10x\\ y'&=\boxed{\boxed{10\cos10x}} \end{aligned}

Jadi, turunannya adalah y'=10\cos10x

No.

Tentukan turunan kedua dari y=2\cos\left(x^3-x\right)

Grup Telegram

\begin{aligned} y&=2\cos\left(x^3-x\right)\\ y'&=2\left(-\sin\left(x^3-x\right)\right)\left(3x^2-1\right)\\ &=\left(-6x^2+2\right)\sin\left(x^3-x\right) \end{aligned}

\begin{aligned} u&=-6x^2+2\\ u'&=-12x \end{aligned}

\begin{aligned} v&=\sin\left(x^3-x\right)\\ v'&=\left(3x^2-1\right)\cos\left(x^3-x\right) \end{aligned}

\begin{aligned} y"&=u'v+uv'\\ &=-12x\sin\left(x^3-x\right)+\left(-6x^2+2\right)\left(3x^2-1\right)\cos\left(x^3-x\right)\\ &=-12x\sin\left(x^3-x\right)-2\left(3x^2-1\right)^2\cos\left(x^3-x\right) \end{aligned}

Jadi, turunan kedua dari y=2\cos\left(x^3-x\right) adalah -12x\sin\left(x^3-x\right)-2\left(3x^2-1\right)^2\cos\left(x^3-x\right).

No.

Gunakan {f'(x)=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}h} untuk menentukan turunan {f(x)=3\sin x} untuk {x=\dfrac{\pi}6}

Grup Telegram

\begin{aligned} f'(x)&=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}h\\ &=\displaystyle\lim_{h\to0}\dfrac{3\sin(x+h)-3\sin x}h\\ &=\displaystyle\lim_{h\to0}\dfrac{3\left(\sin x\cos h+\cos x\sin h\right)-3\sin x}h\\ &=\displaystyle\lim_{h\to0}\dfrac{3\sin x\cos h+3\cos x\sin h-3\sin x}h\\ &=\displaystyle\lim_{h\to0}\dfrac{3\sin x\left(1-2\sin^2\dfrac12h\right)+3\cos x\sin h-3\sin x}h\\ &=\displaystyle\lim_{h\to0}\dfrac{3\sin x-6\sin x\sin^2\dfrac12h+3\cos x\sin h-3\sin x}h\\ &=\displaystyle\lim_{h\to0}\dfrac{-6\sin x\sin^2\dfrac12h+3\cos x\sin h}h\\ &=\displaystyle\lim_{h\to0}\left(-6\sin x\sin\dfrac12h\dfrac{\sin\dfrac12h}h+3\cos x\dfrac{\sin h}h\right)\\ &=-6\sin x\sin\dfrac12(0)\cdot\dfrac12+3\cos x\cdot1\\ &=-6\sin x\sin0\cdot\dfrac12+3\cos x\\ &=-6\sin x(0)\cdot\dfrac12+3\cos x\\ &=3\cos x\\ f'\left(\dfrac{\pi}6\right)&=3\cos\dfrac{\pi}6\\ &=3\left(\dfrac12\sqrt3\right)\\ &=\boxed{\boxed{\dfrac32\sqrt3}} \end{aligned}

Jadi, f'\left(\dfrac{\pi}6\right)=\dfrac32\sqrt3.

No.

y=\sqrt[3]{\sin^2(2x-1)^5}
y'= ...

Grup Telegram

\begin{aligned} y&=\sqrt[3]{\sin^2(2x-1)^5}\\ &=\sin^{\frac23}(2x-1)^5\\ y'&=\dfrac23\sin^{-\frac13}(2x-1)^5\cdot\cos(2x-1)^5\cdot5(2x-1)^4\cdot2\\ &=\dfrac{20(2x-1)^4\cos(2x-1)^5}{3\sin^{\frac13}(2x-1)^5}\\ &=\boxed{\boxed{\dfrac{20(2x-1)^4\cos(2x-1)^5}{3\sqrt[3]{\sin(2x-1)^5}}}} \end{aligned}

Jadi, y'=\dfrac{20(2x-1)^4\cos(2x-1)^5}{3\sqrt[3]{\sin(2x-1)^5}}.

No.

Hitunglah y'(0), jika {y=\dfrac{\sin x}{1+\cos x}}

Grup Telegram

\begin{aligned} u&=\sin x\\ u'&=\cos x\end{aligned}

\begin{aligned} v&=1+\cos x\\ v'&=-\sin x\end{aligned}

\begin{aligned} y&=\dfrac{u}v\\ y'&=\dfrac{u'v-uv'}{v^2}\\ &=\dfrac{\cos x(1+\cos x)-\sin x(-\sin x)}{(1+\cos x)^2}\\ &=\dfrac{\cos x+\cos^2x+\sin^2x}{(1+\cos x)^2}\\ &=\dfrac{\cos x+1}{(1+\cos x)^2}\\ &=\dfrac1{1+\cos x}\\ y'(0)&=\dfrac1{1+\cos0}\\ &=\dfrac1{1+1}\\ &=\boxed{\boxed{\dfrac12}} \end{aligned}

Jadi, y'(0)=\dfrac12.

No.

Hitunglah turunan {y=\sin^3(3x-2)}

Grup Telegram

\begin{aligned} y&=\sin^3(3x-2)\\ y'&=3\sin^2(3x-2)\cdot\cos(3x-2)\cdot3\\ &=\boxed{\boxed{9\sin^2(3x-2)\cos(3x-2)}} \end{aligned}

Jadi, turunan {y=\sin^3(3x-2)} adalah y'=9\sin^2(3x-2)\cos(3x-2).

No.

Titik stasioner dari fungsi {f(x)=\tan^2x} adalah untuk nilai x= ....

Grup Telegram

\begin{aligned} f(x)&=\tan^2x\\ f'(x)&=2\tan x\sec^2x \end{aligned}

f(x) mencapai stasioner saat f'(x)=0
\begin{aligned} 2\tan x\sec^2x&=0\\ \tan x&=0\\ x&=0\degree+k\cdot180\degree\\ &=k\cdot180\degree \end{aligned}

Jadi, titik stasioner dari fungsi {f(x)=\tan^2x} adalah untuk nilai x=k\cdot180\degree.