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)}
   $$\begin{array}{rl}h(x)& ={\displaystyle \frac{{\cos }^{2}x+4\cos x+4}{\cos x+2}}\\ & ={\displaystyle \frac{(\cos x+2{)}^2}{\cos x+2}}\\ & =\cos x+2\\ {\displaystyle \frac{dh(x)}{dx}}& =\boxed{{\displaystyle \boxed{{\displaystyle -\sin x}}}}\end{array}$$
   
   

2. \dfrac{d^2h(x)}{dx^2}=-\cos x
   $$\begin{array}{rl}{\displaystyle \frac{d^{2}h\left({\displaystyle \frac{\pi }{3}}\right)}{d{x}^2}}& =-\cos {\displaystyle \frac{\pi }{3}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{1}{2}}}}}}\end{array}$$

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

$$\begin{array}{rl}y& =x^3\tan 2x-{\displaystyle \frac{1}{2}}\tan 2x+x\tan 2x-\tan 2x\\ & =\tan 2x(x^3-{\displaystyle \frac{1}{2}}{x}^2+x-1)\end{array}$$

$$\begin{array}{rl}u& =\tan 2x\\ u^{\prime }& =2{\sec }^{2}2x\end{array}$$

$$\begin{array}{rl}v& =x^3-{\displaystyle \frac{1}{2}}{x}^2+x-1\\ v^{\prime }& =3x^2-x+1\end{array}$$

$$\begin{array}{rl}{y}^{\prime }& =u^{\prime }v+u{v}^{\prime }\\ & =2{\sec }^{2}2x(x^3-{\displaystyle \frac{1}{2}}{x}^2+x-1)+\tan 2x(3x^2-x+1)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2x^3{\sec }^{2}2x-x^2{\sec }^{2}2x+2x{\sec }^{2}2x-2{\sec }^{2}2x+3x^2\tan 2x-x\tan 2x+\tan 2x}}}}\end{array}$$

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

$$\begin{array}{rl}y& =\sin 10x\\ y^{\prime }& =\boxed{{\displaystyle \boxed{{\displaystyle 10\cos 10x}}}}\end{array}$$

Jadi, turunannya adalah y'=10\cos10x

No.

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

Grup Telegram

$$\begin{array}{rl}y& =2\cos (x^3-x)\\ y^{\prime }& =2(-\sin (x^3-x))(3x^2-1)\\ & =(-6x^2+2)\sin (x^3-x)\end{array}$$

$$\begin{array}{rl}u& =-6x^2+2\\ u^{\prime }& =-12x\end{array}$$

$$\begin{array}{rl}v& =\sin (x^3-x)\\ v^{\prime }& =(3x^2-1)\cos (x^3-x)\end{array}$$

$$\begin{array}{rl}y"& =u^{\prime }v+u{v}^{\prime }\\ & =-12x\sin (x^3-x)+(-6x^2+2)(3x^2-1)\cos (x^3-x)\\ & =-12x\sin (x^3-x)-2{(3x^2-1)}^2\cos (x^3-x)\end{array}$$

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

$$\begin{array}{rl}{f}^{\prime }(x)& ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)-f(x)}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{3\sin (x+h)-3\sin x}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{3(\sin x\cos h+\cos x\sin h)-3\sin x}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{3\sin x\cos h+3\cos x\sin h-3\sin x}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{3\sin x(1-2{\sin }^2{\displaystyle \frac{1}{2}}h)+3\cos x\sin h-3\sin x}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{3\sin x-6\sin x{\sin }^2{\displaystyle \frac{1}{2}}h+3\cos x\sin h-3\sin x}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{-6\sin x{\sin }^2{\displaystyle \frac{1}{2}}h+3\cos x\sin h}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}(-6\sin x\sin {\displaystyle \frac{1}{2}}h{\displaystyle \frac{\sin {\displaystyle \frac{1}{2}}h}{h}}+3\cos x{\displaystyle \frac{\sin h}{h}})}\\ & =-6\sin x\sin {\displaystyle \frac{1}{2}}(0)\cdot {\displaystyle \frac{1}{2}}+3\cos x\cdot 1\\ & =-6\sin x\sin 0\cdot {\displaystyle \frac{1}{2}}+3\cos x\\ & =-6\sin x(0)\cdot {\displaystyle \frac{1}{2}}+3\cos x\\ & =3\cos x\\ f^{\prime }\left({\displaystyle \frac{\pi }{6}}\right)& =3\cos {\displaystyle \frac{\pi }{6}}\\ & =3\left({\displaystyle \frac{1}{2}}\sqrt{3}\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{3}{2}}\sqrt{3}}}}}\end{array}$$

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

No.

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

Grup Telegram

$$\begin{array}{rl}y& =\sqrt[3]{{\sin }^2(2x-1{)}^5}\\ & ={\sin }^{\frac{2}{3}}(2x-1{)}^5\\ y^{\prime }& ={\displaystyle \frac{2}{3}}{\sin }^{-\frac{1}{3}}(2x-1{)}^5\cdot \cos (2x-1{)}^5\cdot 5(2x-1{)}^4\cdot 2\\ & ={\displaystyle \frac{20(2x-1{)}^4\cos (2x-1{)}^5}{3{\sin }^{\frac{1}{3}}(2x-1{)}^5}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{20(2x-1{)}^4\cos (2x-1{)}^5}{3\sqrt[3]{\sin (2x-1{)}^5}}}}}}}\end{array}$$

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

$$\begin{array}{rl}u& =\sin x\\ u^{\prime }& =\cos x\end{array}$$

$$\begin{array}{rl}v& =1+\cos x\\ v^{\prime }& =-\sin x\end{array}$$

$$\begin{array}{rl}y& ={\displaystyle \frac{u}{v}}\\ y^{\prime }& ={\displaystyle \frac{u^{\prime }v-u{v}^{\prime }}{v^2}}\\ & ={\displaystyle \frac{\cos x(1+\cos x)-\sin x(-\sin x)}{(1+\cos x{)}^2}}\\ & ={\displaystyle \frac{\cos x+{\cos }^{2}x+{\sin }^{2}x}{(1+\cos x{)}^2}}\\ & ={\displaystyle \frac{\cos x+1}{(1+\cos x{)}^2}}\\ & ={\displaystyle \frac{1}{1+\cos x}}\\ y^{\prime }(0)& ={\displaystyle \frac{1}{1+\cos 0}}\\ & ={\displaystyle \frac{1}{1+1}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{2}}}}}}\end{array}$$

Jadi, y'(0)=\dfrac12.

No.

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

Grup Telegram

$$\begin{array}{rl}y& ={\sin }^3(3x-2)\\ y^{\prime }& =3{\sin }^2(3x-2)\cdot \cos (3x-2)\cdot 3\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 9{\sin }^2(3x-2)\cos (3x-2)}}}}\end{array}$$

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

$$\begin{array}{rl}f(x)& ={\tan }^{2}x\\ f^{\prime }(x)& =2\tan x{\sec }^{2}x\end{array}$$

f(x) mencapai stasioner saat f'(x)=0
$$\begin{array}{rl}2\tan x{\sec }^{2}x& =0\\ \tan x& =0\\ x& =0{}^{\circ }+k\cdot 180{}^{\circ }\\ & =k\cdot 180{}^{\circ }\end{array}$$

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