Exercise Zone: Turunan Fungsi Trigonometri [2]
Table of Contents
Tipe:
No.
Diketahui- Tentukan turunan pertama dari h(x), jika diketahui \(h(x)=\dfrac{f(x)}{g(x)}\)
- Hitunglah nilai dari \(\dfrac{d^2h(x)}{dx^2}\) untuk \(x=\dfrac{\pi}3\)
ALTERNATIF PENYELESAIAN
- \(h(x)=\dfrac{f(x)}{g(x)}\)
\(\begin{aligned} h(x)&=\dfrac{\cos^2x+4\cos x+4}{\cos x+2}\\[4pt] &=\dfrac{(\cos x+2)^2}{\cos x+2}\\[4pt] &=\cos x+2\\ \dfrac{dh(x)}{dx}&=\boxed{\boxed{-\sin x}} \end{aligned}\)
- \(\dfrac{d^2h(x)}{dx^2}=-\cos x\)
\(\begin{aligned} \dfrac{d^2h\left(\dfrac{\pi}3\right)}{dx^2}&=-\cos\dfrac{\pi}3\\ &=\boxed{\boxed{-\dfrac12}} \end{aligned}\)
Jadi,
JAWAB:
- turunan pertama dari h(x), jika diketahui \(h(x)=\dfrac{f(x)}{g(x)}\) adalah
−sin x - nilai dari \(\dfrac{d^2h(x)}{dx^2}\) untuk \(x=\dfrac{\pi}3\) adalah \(-\dfrac12\)
JAWAB:
No.
Tentukan y' jlka diketahui fungsi y sebagai berikut! \[y=x^3\tan2x-\dfrac12x^2\tan2x+x\tan2x-\tan2x\]ALTERNATIF PENYELESAIAN
\(\begin{aligned}
y&=x^3\tan2x-\dfrac12\tan2x+x\tan2x-\tan2x\\
&=\tan2x\left(x^3-\dfrac12x^2+x-1\right)
\end{aligned}\)
\(\begin{aligned} y'&=u'v+uv'\\ &=2\sec^22x\left(x^3-\dfrac12x^2+x-1\right)+\tan2x\left(3x^2-x+1\right)\\ &=\boxed{\boxed{2x^3\sec^22x-x^2\sec^22x+2x\sec^22x-2\sec^22x+3x^2\tan2x-x\tan2x+\tan2x}} \end{aligned}\)
\(\begin{aligned}
u&=\tan2x\\
u'&=2\sec^22x
\end{aligned}\)
\(\begin{aligned}
v&=x^3-\dfrac12x^2+x-1\\
v'&=3x^2-x+1
\end{aligned}\)
\(\begin{aligned} y'&=u'v+uv'\\ &=2\sec^22x\left(x^3-\dfrac12x^2+x-1\right)+\tan2x\left(3x^2-x+1\right)\\ &=\boxed{\boxed{2x^3\sec^22x-x^2\sec^22x+2x\sec^22x-2\sec^22x+3x^2\tan2x-x\tan2x+\tan2x}} \end{aligned}\)
Jadi, y' = 2x^3 sec^22x - x^2\sec^22x + 2x sec^22x -2 sec^22x + 3x^2\tan 2x -x tan 2x + tan 2x .
No.
Tentukan turunan dari fungsi trigonometri berikut!y = sin 10x
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
y&=\sin10x\\
y'&=\boxed{\boxed{10\cos10x}}
\end{aligned}\)
Jadi, turunannya adalah y' = 10 cos 10x .
No.
Tentukan turunan kedua dariALTERNATIF PENYELESAIAN
\(\begin{aligned}
y&=2\cos\left(x^3-x\right)\\
y'&=2\left(-\sin\left(x^3-x\right)\right)\left(3x^2-1\right)\\
&=\left(-6x^2+2\right)\sin\left(x^3-x\right)
\end{aligned}\)
\(\begin{aligned} y"&=u'v+uv'\\ &=-12x\sin\left(x^3-x\right)+\left(-6x^2+2\right)\left(3x^2-1\right)\cos\left(x^3-x\right)\\ &=-12x\sin\left(x^3-x\right)-2\left(3x^2-1\right)^2\cos\left(x^3-x\right) \end{aligned}\)
\(\begin{aligned}
u&=-6x^2+2\\
u'&=-12x
\end{aligned}\)
\(\begin{aligned}
v&=\sin\left(x^3-x\right)\\
v'&=\left(3x^2-1\right)\cos\left(x^3-x\right)
\end{aligned}\)
\(\begin{aligned} y"&=u'v+uv'\\ &=-12x\sin\left(x^3-x\right)+\left(-6x^2+2\right)\left(3x^2-1\right)\cos\left(x^3-x\right)\\ &=-12x\sin\left(x^3-x\right)-2\left(3x^2-1\right)^2\cos\left(x^3-x\right) \end{aligned}\)
Jadi, turunan kedua dari y = 2 cos(x3 − x) adalah −12x sin(x3 − x) − 2(3x2 − 1)2 cos(x3 − x) .
No.
Gunakan \(f'(x)=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}h\) untuk menentukan turunanALTERNATIF PENYELESAIAN
\(\begin{aligned}
f'(x)&=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}h\\
&=\displaystyle\lim_{h\to0}\dfrac{3\sin(x+h)-3\sin x}h\\
&=\displaystyle\lim_{h\to0}\dfrac{3\left(\sin x\cos h+\cos x\sin h\right)-3\sin x}h\\
&=\displaystyle\lim_{h\to0}\dfrac{3\sin x\cos h+3\cos x\sin h-3\sin x}h\\
&=\displaystyle\lim_{h\to0}\dfrac{3\sin x\left(1-2\sin^2\dfrac12h\right)+3\cos x\sin h-3\sin x}h\\
&=\displaystyle\lim_{h\to0}\dfrac{3\sin x-6\sin x\sin^2\dfrac12h+3\cos x\sin h-3\sin x}h\\
&=\displaystyle\lim_{h\to0}\dfrac{-6\sin x\sin^2\dfrac12h+3\cos x\sin h}h\\
&=\displaystyle\lim_{h\to0}\left(-6\sin x\sin\dfrac12h\dfrac{\sin\dfrac12h}h+3\cos x\dfrac{\sin h}h\right)\\
&=-6\sin x\sin\dfrac12(0)\cdot\dfrac12+3\cos x\cdot1\\
&=-6\sin x\sin0\cdot\dfrac12+3\cos x\\
&=-6\sin x(0)\cdot\dfrac12+3\cos x\\
&=3\cos x\\
f'\left(\dfrac{\pi}6\right)&=3\cos\dfrac{\pi}6\\
&=3\left(\dfrac12\sqrt3\right)\\
&=\boxed{\boxed{\dfrac32\sqrt3}}
\end{aligned}\)
Jadi, \(f'\left(\dfrac{\pi}6\right)=\dfrac32\sqrt3\).
No.
\(y=\sqrt[3]{\sin^2(2x-1)^5}\)y' = ...
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
y&=\sqrt[3]{\sin^2(2x-1)^5}\\
&=\sin^{\frac23}(2x-1)^5\\
y'&=\dfrac23\sin^{-\frac13}(2x-1)^5\cdot\cos(2x-1)^5\cdot5(2x-1)^4\cdot2\\
&=\dfrac{20(2x-1)^4\cos(2x-1)^5}{3\sin^{\frac13}(2x-1)^5}\\
&=\boxed{\boxed{\dfrac{20(2x-1)^4\cos(2x-1)^5}{3\sqrt[3]{\sin(2x-1)^5}}}}
\end{aligned}\)
Jadi, \(y'=\dfrac{20(2x-1)^4\cos(2x-1)^5}{3\sqrt[3]{\sin(2x-1)^5}}\).
No.
Hitunglah y'(0), jika \(y=\dfrac{\sin x}{1+\cos x}\)ALTERNATIF PENYELESAIAN
\(\begin{aligned} u&=\sin x\\
u'&=\cos x\end{aligned}\)
\(\begin{aligned} v&=1+\cos x\\
v'&=-\sin x\end{aligned}\)
\(\begin{aligned} y&=\dfrac{u}v\\ y'&=\dfrac{u'v-uv'}{v^2}\\[4pt] &=\dfrac{\cos x(1+\cos x)-\sin x(-\sin x)}{(1+\cos x)^2}\\[4pt] &=\dfrac{\cos x+\cos^2x+\sin^2x}{(1+\cos x)^2}\\[4pt] &=\dfrac{\cos x+1}{(1+\cos x)^2}\\[4pt] &=\dfrac1{1+\cos x}\\[4pt] y'(0)&=\dfrac1{1+\cos0}\\[4pt] &=\dfrac1{1+1}\\ &=\boxed{\boxed{\dfrac12}} \end{aligned}\)
Jadi, \(y'(0)=\dfrac12\).
No.
Hitunglah turunanALTERNATIF PENYELESAIAN
\(\begin{aligned} y&=\sin^3(3x-2)\\
y'&=3\sin^2(3x-2)\cdot\cos(3x-2)\cdot3\\
&=\boxed{\boxed{9\sin^2(3x-2)\cos(3x-2)}}
\end{aligned}\)
Jadi, turunan y = sin3(3x − 2) adalah y' = 9 sin2(3x − 2) cos(3x − 2) .
No.
Titik stasioner dari fungsiALTERNATIF PENYELESAIAN
\(\begin{aligned}
f(x)&=\tan^2x\\
f'(x)&=2\tan x\sec^2x
\end{aligned}
\)
f(x) mencapai stasioner saatf'(x) = 0
\(\begin{aligned} 2\tan x\sec^2x&=0\\ \tan x&=0\\ x&=0\degree+k\cdot180\degree\\ &=k\cdot180\degree \end{aligned} \)
f(x) mencapai stasioner saat
\(\begin{aligned} 2\tan x\sec^2x&=0\\ \tan x&=0\\ x&=0\degree+k\cdot180\degree\\ &=k\cdot180\degree \end{aligned} \)
Jadi, titik stasioner dari fungsi f(x) = tan2 x adalah untuk nilai x = k ⋅ 180°
No.
Turunan pertama dari fungsi \(f(x)=\dfrac{2x+4}{\sin x}\) adalah ....- \(\dfrac{2\left(\sin x-x\cos x-2\cos x\right)}{\sin^2x}\)
- \(\dfrac{\sin x-x\cos x+2\cos x}{\sin^2x}\)
- \(\dfrac{2\sin x-x\cos x+2\cos x}{\sin^2x}\)
- \(\dfrac{2\left(\sin x-\cos x\right)}{\sin x}\)
- \(\dfrac{2\left(\sin x+\cos x\right)}{\sin x}\)
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
u&=2x+4\\
u'&=2
\end{aligned}\)
\(\begin{aligned}
v&=\sin x\\
v'&=\cos x
\end{aligned}\)
\(\begin{aligned} f'(x)&=\dfrac{u'v-uv'}{v^2}\\[4pt] &=\dfrac{2\sin x-(2x+4)\cos x}{(\sin x)^2}\\[4pt] &=\dfrac{2\sin x-2x\cos x-4\cos x}{\sin^2 x}\\ &=\boxed{\boxed{\dfrac{2\left(\sin x-x\cos x-2\cos x\right)}{\sin^2x}}} \end{aligned}\)
Jadi, turunan pertama dari fungsi \(f(x)=\dfrac{2x+4}{\sin x}\) adalah \(\dfrac{2\left(\sin x-x\cos x-2\cos x\right)}{\sin^2x}\).
JAWAB: A
JAWAB: A
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