VISUALISASI TURUNAN FUNGSI COSINUS

Visualisasi Turunan Kosinus: $f(x) = \cos(x) \implies f'(x) = -\sin(x)$

Eksplorasi hubungan gradien garis singgung $f(x)$ dengan nilai fungsi turunan $f'(x)$.

Turunan Kosinus $\frac{d}{dx}\cos(x) = -\sin(x)$

$f(x) = \cos(x)$

Garis Singgung ($m$)

$f'(x) = -\sin(x)$

Grafik Utama: $f(x) = \cos(x)$

Grafik Turunan: $f'(x) = -\sin(x)$

Kecepatan: 0.5x 1.0x 1.5x 2.0x

-2π 2π

Garis Singgung (Tangent Line) Segitiga Gradien ($\Delta y / \Delta x$) Jejak Kurva $f'(x)$

Data Parameter Real-Time

Posisi $x$

0.00 rad

0°

$f(x) = \cos(x)$ 1.00

Gradien ($m$) 0.00

$f'(x) = -\sin(x)$ 0.00

Intuisi Matematika

Geser slider atau tekan putar untuk melihat hubungan antara kemiringan garis singgung $f(x)$ dengan nilai turunan $f'(x)$.

Pasang di Blog Anda

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Kode iFrame berhasil disalin!

VISUALISASI TURUNAN FUNGSI SINUS

Visualisasi Turunan: f(x) = sin(x) → f'(x) = cos(x)

Gradien garis singgung (kemiringan) pada kurva f(x) = sin(x) di titik mana pun selalu sama dengan nilai fungsi f'(x) = cos(x).

f(x) = sin(x)

Garis Singgung (Gradien m)

f'(x) = cos(x)

Posisi x 0.00 rad

-2π 0 2π

Nilai x

0.00

0.00 π rad

sin(x)

0.000

Tinggi Kurva

Gradien (m)

1.000

Kemiringan

cos(x)

1.000

Hasil Turunan

💡 Cara Membaca Visualisasi:

• Garis putus-putus putih menghubungkan titik pada sin(x) dengan titik turunannya pada cos(x).
• Perhatikan kemiringan (Gradien m) dari garis singgung emas. Nilai m ini selalu persis sama dengan tinggi kurva hijau cos(x).
• Saat kurva sinus berada di puncak atau lembah, garis singgung mendatar (gradien $m = 0$), dan nilai cosinus berada persis di angka $0$.

VISUALISASI TURUNAN FUNGSI ALJABAR

Visualisasi Interaktif Turunan Fungsi

Geser slider untuk melihat bagaimana kemiringan garis singgung ($f'(x)$) berubah.

Posisi Titik $x$: x = 1.00

Informasi Real-time:

• Fungsi: $f(x) = x^3 - 3x$
• Nilai Titik: $(x, y) =$ (1.00, -2.00)
• Turunan $f'(x) = 3x^2 - 3$: 0.00 (Kemiringan Garis Singgung)

CONTOH SOAL 8 TURUNAN PERTAMA FUNGSI TRIGONOMETRI

 

$\begin{array}{ll}\\ 36.&\textrm{Diketahui}\: \: f(x)=\displaystyle \frac{\sin x-\cos x}{\tan x}. \: \: \textrm{Nilai}\\ &\textrm{turunan pertama fungsi}\: \: f\: \: \textrm{saat}\: \: x=45^{\circ}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\sqrt{2}\\ \textrm{b}.&\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{c}.&\displaystyle 1\\ \textrm{d}.&\displaystyle \sqrt{2}\\ \textrm{e}.&\displaystyle \sqrt{3} \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{\sin x-\cos x}{\tan x}\\ \textrm{Guna}&\textrm{kan formula}\: \: y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\sin x-\cos x \Rightarrow u'=\cos x+\sin x\\ v&=\tan x\Rightarrow v'=\sec ^{2}x\\ \textrm{maka}&\\ f'(x)&=\displaystyle \frac{(\cos x+\sin x).\tan x-(\sin x-\cos x).\sec ^{2}x}{\tan ^{2}x}\\ f'\left ( 45^{\circ} \right )&=\displaystyle \frac{\left ( \displaystyle \frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2} \right ).1-\left ( \displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2} \right ).\left ( \sqrt{2} \right )^{2}}{1^{2}}\\ &=\displaystyle \frac{\sqrt{2}-0}{1}\\ &=\sqrt{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 37.&\textrm{Turunan pertama dari fungsi}\\ &g(x)=\displaystyle \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{\cos ^{2}x}-\frac{1}{\sin ^{2}x}\\ \textrm{b}.&\displaystyle \frac{1}{\cos ^{2}x}+\frac{1}{\sin ^{2}x}\\ \textrm{c}.&\displaystyle \frac{1}{\sin^{2} x\cos ^{2}x}\\ \textrm{d}.&\displaystyle \frac{-1}{\sin ^{2}x\cos ^{2}x}\\ \textrm{e}.&\displaystyle \sin ^{2}x\cos ^{2}x \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ g(x)&=\displaystyle \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\\ &=\frac{\sin ^{2}x+\cos ^{2}x}{\sin x\cos x}=\displaystyle \frac{1}{\sin x\cos x}\\ \textrm{maka}&\\ g'(x)&=\displaystyle \frac{0.(\sin x\cos x)-1.\left (\cos ^{2}x -\sin ^{2}x \right )}{(\sin x\cos x)^{2}}\\ &=\displaystyle \frac{\sin ^{2}x-\cos ^{2}x}{\sin^{2} x\cos^{2} x}\\ &=\displaystyle \frac{1}{\cos ^{2}x}-\frac{1}{\sin ^{2}x} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 38.&\textrm{Diketahui}\: \: h(x)=\cos \left ( \displaystyle \frac{3}{x} \right ), \\ &\textrm{maka}\: \: \displaystyle \frac{dh}{dx}\\ &\begin{array}{llll}\\ \textrm{a}.&-3\sin \displaystyle \frac{3}{x}\\ \textrm{b}.&-\displaystyle \frac{3}{x^{2}}\sin \frac{3}{x}\\ \textrm{c}.&-\displaystyle \frac{3}{x}\sin \frac{3}{x}\\ \textrm{d}.&\displaystyle \frac{3}{x^{2}}\sin \frac{3}{x}\\ \textrm{e}.&\displaystyle \frac{3}{x}\sin \frac{3}{x} \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\cos \displaystyle \frac{3}{x}&=-\sin \displaystyle \frac{3}{x}\left ( \displaystyle \frac{0.(x)-3.1}{x^{2}} \right )\\ &=\displaystyle \frac{-(-3)}{x^{2}}\sin \frac{3}{x}\\ &=\displaystyle \frac{3}{x^{2}}\sin \frac{3}{x} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 39.&\textrm{Turunan pertama dari}\: \: \tan (\cos x), \\ &\textrm{terhadap}\: \: x\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&-\sec ^{2}(\cos x)\sin x\\ \textrm{b}.&\sec ^{2}(\cos x)\sin x\\ \textrm{c}.&\sec ^{2}(\sin x)\cos x\\ \textrm{d}.&\displaystyle \sin x\\ \textrm{e}.&\displaystyle -\sin x \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Misal}&\textrm{kan}\\ y&=\tan x(\cos x)\\ y'&=\sec ^{2}(\cos x)\times (-\sin x)\\ &=-\sec ^{2}(\cos x).\sin x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 40.&(\textbf{UN 2005})\textrm{Turunan pertama dari}\\ &f(x)=\sqrt[3]{\cos ^{2}\left ( 3x^{2}+5x \right )}\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{2}{3}\cos ^{.^{-\frac{1}{3}}}\left ( 3x^{2}+5x \right )\sin \left ( 3x^{2}+5x \right )\\ \textrm{b}.&\displaystyle \frac{2}{3}(6x+5)\cos ^{.^{-\frac{1}{3}}}\left ( 3x^{2}+5x \right )\\ \textrm{c}.&-\displaystyle \frac{2}{3}\cos^{.^{-\frac{1}{3}}} \left ( 3x^{2}+5x \right )\sin \left ( 3x^{2}+5x \right )\\ \textrm{d}.&-\displaystyle \frac{2}{3}(6x+5)\tan \left ( 3x^{2}+5x \right )\sqrt[3]{\cos ^{2}\left ( 3x^{2}+5x \right )}\\ \textrm{e}.&\displaystyle \frac{2}{3}(6x+5)\tan \left ( 3x^{2}+5x \right )\sqrt[3]{\cos ^{2}\left ( 3x^{2}+5x \right )} \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\textrm{Misal}&\textrm{kan}\\ f(x)&=\sqrt[3]{\cos ^{2}\left ( 3x^{2}+5x \right )}\\ f'(x)&=\cos ^{.^{\frac{2}{3}}}\left ( 3x^{2}+5x \right )\\ &=\displaystyle \frac{2}{3}\cos ^{.^{-\frac{1}{2}}}\left ( 3x^{2}+5x \right )\times \left ( -\sin \left ( 3x^{2}+5x \right ) \right )\\ &\qquad\qquad\qquad\qquad \times (6x+5)\\ &=-\displaystyle \frac{2}{3}(6x+5)\cos ^{.^{-\frac{1}{3}}}\left ( 3x^{2}+5x \right )\sin \left ( 3x^{2}+5x \right )\\ &=-\displaystyle \frac{2}{3}(6x+5)\cos ^{.^{\frac{2}{3}}}\left ( 3x^{2}+5x \right )\\ &\times \cos^{-1} \left ( 3x^{2}+5x \right )\times \sin \left ( 3x^{2}+5x \right )\\ &=-\displaystyle \frac{2}{3}(6x+5)\tan \left ( 3x^{2}+5x \right )\sqrt[3]{\cos ^{2}\left ( 3x^{2}+5x \right )} \end{aligned} \end{array}$

CONTOH SOAL 7 TURUNAN PERTAMA FUNGSI TRIGONOMETRI

 

$\begin{array}{ll}\\ 31.&\textrm{Turunan pertama dari}\: \: f(x)=\displaystyle \frac{\sin x}{x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{x\cos x+\sin x}{x^{2}}\\ \textrm{b}.&\displaystyle \frac{x\cos x-\sin x}{x^{2}}\\ \textrm{c}.&\displaystyle \frac{-x\cos x-\sin x}{x^{2}}\\ \textrm{d}.&\displaystyle \frac{\cos x-x\sin x}{x^{2}}\\ \textrm{e}.&\displaystyle \frac{\cos x+x\sin x}{x^{2}} \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{\sin x}{x}\\ \textrm{Guna}&\textrm{kan formula}\: \: y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\sin x\Rightarrow u'=\cos x\\ v&=x\Rightarrow v'=1\\ \textrm{maka}&\\ f'(x)&=\displaystyle \frac{\cos x.(x)-\sin x.1}{x^{2}}\\ &=\displaystyle \frac{x\cos x-\sin x}{x^{2}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 32.&\textrm{Turunan pertama dari}\: \: f(x)=\displaystyle \frac{1-\cos x}{x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{x\sin x+\cos x+1}{x^{2}}\\ \textrm{b}.&\displaystyle \frac{x\cos x+\sin x-1}{x^{2}}\\ \textrm{c}.&\displaystyle \frac{x\sin x-\cos x+1}{x^{2}}\\ \textrm{d}.&\displaystyle \frac{x\sin x+\cos x-1}{x^{2}}\\ \textrm{e}.&\displaystyle \frac{x\cos x-\sin x+1}{x^{2}} \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{1-\cos x}{x}\\ \textrm{Guna}&\textrm{kan formula}\: \: y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=1-\cos x\Rightarrow u'=\sin x\\ v&=x\Rightarrow v'=1\\ \textrm{maka}&\\ f'(x)&=\displaystyle \frac{\sin x.(x)-(1-\cos x).1}{x^{2}}\\ &=\displaystyle \frac{x\sin x+\cos x-1}{x^{2}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 33.&\textrm{Turunan pertama dari}\: \: f(x)=\displaystyle \frac{\tan x}{\cos x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1+\cos ^{2}x}{\cos ^{3}x}\\ \textrm{b}.&\displaystyle \frac{1-\cos x}{\cos ^{3}x}\\ \textrm{c}.&\displaystyle \frac{1+\sin ^{2}x}{\cos ^{3}x}\\ \textrm{d}.&\displaystyle \frac{1+\sin x}{\cos ^{3}x}\\ \textrm{e}.&\displaystyle \frac{1-\sin ^{2}x}{\cos ^{3}x} \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{\tan x}{\cos x}\\ \textrm{Guna}&\textrm{kan formula}\: \: y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\tan x\Rightarrow u'=\sec ^{2}x\\ v&=\cos x\Rightarrow v'=-\sin x\\ \textrm{maka}&\\ f'(x)&=\displaystyle \frac{\sec ^{2}x.(\cos x)-(\tan x).(-\sin x)}{\cos ^{2}x}\\ &=\displaystyle \frac{\sec ^{2}x.\cos x+\tan x\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{\left ( \displaystyle \frac{1}{\cos ^{2}x} \right )\cos x+\left ( \displaystyle \frac{\sin x}{\cos x} \right )\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{\displaystyle \frac{1}{\cos x}+\displaystyle \frac{\sin ^{2}x}{\cos x}}{\cos ^{2}x}\\ &=\displaystyle \frac{1+\sin ^{2}x}{\cos ^{3}x} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 34.&\textrm{Turunan pertama dari}\: \: g(t)=\displaystyle \frac{\cos t+2t}{\sin t} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{2\sin t+2t\cos t-1}{\sin^{2} t}\\ \textrm{b}.&\displaystyle \frac{2\sin t-2t\cos t+1}{\sin^{2} t}\\ \textrm{c}.&\displaystyle \frac{2\sin t+2t\cos t+1}{\sin^{2} t}\\ \textrm{d}.&\displaystyle \frac{2\sin t-2t\cos t-1}{\sin^{2} t}\\ \textrm{e}.&\displaystyle \frac{-2\sin t+2t\cos t-1}{\sin^{2} t} \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ g(t)&=\displaystyle \frac{\cos t+2t}{\sin t}\\ \textrm{Guna}&\textrm{kan formula}\: \: y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\cos t+2t\Rightarrow u'=-\sin t+2\\ v&=\sin t\Rightarrow v'=\cos t\\ \textrm{maka}&\\ g'(t)&=\displaystyle \frac{(-\sin t+2)(\sin t)-(\cos t+2t)(\cos t)}{\sin ^{2}t}\\ &=\displaystyle \frac{-\sin ^{2}t+2\sin t-\cos ^{2}t-2t\cos t}{\sin ^{2}t}\\ &=\displaystyle \frac{t+2\sin t-2t\cos t-\sin ^{2}t-\cos ^{2}t}{\sin ^{2}t}\\ &=\displaystyle \frac{t+2\sin t-2t\cos t-\left (\sin ^{2}t+\cos ^{2}t \right )}{\sin ^{2}t}\\ &=\displaystyle \frac{2\sin t-2t\cos t-1}{\sin ^{2}t} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 35.&\textrm{Turunan pertama dari}\: \: h(x)=\displaystyle \frac{\sin x}{\sin x+\cos x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{\cos ^{2}x-\sin ^{2}x}\\ \textrm{b}.&\displaystyle \frac{1}{\sin ^{2}x-\cos ^{2}x}\\ \textrm{c}.&\displaystyle \frac{1}{(\sin x+\cos x)^{2}}\\ \textrm{d}.&\displaystyle \sin ^{2}x-\cos ^{2}x\\ \textrm{e}.&\displaystyle 1 \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ h(x)&=\displaystyle \frac{\sin x}{\sin x+\cos x}\\ \textrm{Guna}&\textrm{kan formula}\: \: y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\sin x \Rightarrow u'=\cos x\\ v&=\sin x+\cos x\Rightarrow v'=\cos x-\sin x\\ \textrm{maka}&\\ h'(x)&=\displaystyle \frac{\cos x.(\sin x+\cos x)-\sin x.(\cos x-\sin x)}{(\sin x+\cos x)^{2}}\\ &=\displaystyle \frac{\cos x\sin x+\cos ^{2}x-\sin x\cos x+\sin ^{2}x}{(\sin x+\cos x)^{2}}\\ &=\displaystyle \frac{\sin ^{2}x+\cos ^{2}x}{(\sin x+\cos x)^{2}}\\ &=\displaystyle \frac{1}{(\sin x+\cos x)^{2}} \end{aligned} \end{array}$

CONTOH SOAL 6 TURUNAN PERTAMA FUNGSI TRIGONOMETRI

 

$\begin{array}{ll}\\ 26.&\textrm{Turunan pertama fungsi}\: \: f(x)=\sqrt{\sin x},\\ &\textrm{adalah}\: \: f'(x)=....\\ &\begin{array}{llll} \textrm{a}.&\displaystyle \frac{1}{2\sqrt{\sin x}} \\ \textrm{b}.&\displaystyle \frac{\cos x}{\sqrt{\sin x}}\\ \textrm{c}.&\displaystyle \frac{\cos x}{2\sqrt{\sin x}}\\ \textrm{d}.&-\displaystyle \frac{\sin x}{2\sqrt{\cos x}}\\ \textrm{e}.&\displaystyle \frac{2\cos x}{\sqrt{\sin x}} \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll}&\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\: f(x)=\sqrt{\sin x}=\sin ^{.^{\frac{1}{2}}}x\\ f'(x)&=\displaystyle \frac{1}{2}\left ( \sin ^{.^{-\frac{1}{2}}}x \right ).(\cos x)\\ &=\displaystyle \frac{\cos x}{2\sin ^{.^{\frac{1}{2}}}x}\\ &=\displaystyle \frac{\cos x}{2\sqrt{\sin x}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&\textrm{Jika}\: \: g'(x)\: \: \textrm{adalah turunan pertama}\\ &\textrm{fungsi}\: \: g(x)\: \: \textrm{dengan}\: \: g(x)=5\tan ^{2}x,\\ &\textrm{maka}\: \: g'(x)=....\\ &\begin{array}{llll} \textrm{a}.&10\cos ^{2}x\sin x\\ \textrm{b}.&10\sin ^{2}x\cos x\\ \textrm{c}.&\displaystyle \frac{10\sin x}{\cos ^{3}x}\\ \textrm{d}.&\displaystyle \frac{10\cos ^{3}x}{\sin x}\\ \textrm{e}.&\displaystyle \frac{10}{\sin ^{2}x-\cos ^{2}x} \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \color{black}g(x)=5\tan ^{2}x\\ g'(x)&=5\left ( 2\tan x \right ).\left ( \sec ^{2}x \right )\\ &=10\tan x\times \left ( \displaystyle \frac{1}{\cos ^{2}x} \right )\\ &=10\left ( \displaystyle \frac{\sin x}{\cos x} \right )\times \left ( \displaystyle \frac{1}{\cos ^{2}x} \right )\\ &=\displaystyle \frac{10\sin x}{\cos ^{3}x} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&\textrm{Turunan pertama fungsi}\\ &h(x)=5\sin x\cos x\: \: \textrm{adalah}\: \: h'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&5\sin 2x\\ \textrm{b}.&5\cos 2x\\ \textrm{c}.&5\sin ^{2}x\cos x\\ \textrm{d}.&5\sin ^{2}x\cos^{2} x\\ \textrm{e}.&5\sin 2x\cos x \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll}\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \color{black}h(x)=5\sin x\cos x\\ h(x)&=\displaystyle \frac{5}{2}\left ( 2\sin x\cos x \right )=\displaystyle \frac{5}{2}\sin 2x\\ h'(x)&=\displaystyle \frac{5}{2}\left ( \cos 2x \right ).(2)\\ &=5\cos 2x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 29.&\textrm{Jika diketahui}\: \: f(x)=\left | \tan x \right |,\: \textrm{maka}\: \displaystyle \frac{dy}{dx}\: \textrm{saat}\\ &x=k\:\:\: \text{di mana}\:\:\: \displaystyle \frac{1}{2}\pi\lt k\lt \pi\:\:\: \textrm{adalah}\: ....\\ &\begin{array}{llll} \textrm{a}.&-\sin k\\ \textrm{b}.&\cos k\\ \textrm{c}.&-\sec ^{2}k\\ \textrm{d}.&\sec ^{2}k\\ \textrm{e}.&\cot k \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \: f(x)=\left |\tan x \right |\\ \textrm{saat}&\: \: x=k\: \: \textrm{dengan}\: \: \displaystyle \frac{1}{2}\pi\lt k\lt \pi\\ \textrm{adal}&\textrm{ah}:\\ f(x)&=\left | \tan x \right |,\: \: \textrm{maka saat}\: \: x=k\\ f(k)&=\left | \tan k \right |=-\tan k,\: \: \textrm{karena di}\: \: \displaystyle \frac{1}{2}\pi\lt k\lt \pi\\ \displaystyle \frac{dy}{dx}&=f'(k)=-\sec ^{2}k \end{aligned} \end{array}$

$\begin{array}{ll}\\ 30.&\textrm{Turunan pertama}\: \: g(x)=\left | \cos x \right |\\ & \textrm{adalah}\: \: g'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&-\left | \sin x \right |\\ \textrm{b}.&-\sin x\\ \textrm{c}.&\displaystyle \frac{\sin 2x}{2\left | \cos x \right |}\\ \textrm{d}.&-\displaystyle \frac{\sin 2x}{2\left | \cos x \right |}\\ \textrm{e}.&\left | \sin x \right | \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \: g(x)=\left |\cos x \right |=\sqrt{\cos ^{2}x}=\left ( \cos ^{2}x \right )^{.^{\frac{1}{2}}}\\ g'(x)&=\displaystyle \frac{1}{2}\left ( \cos ^{2}x \right )^{.^{-\frac{1}{2}}}.\left ( 2\cos x \right ).\left ( -\sin x \right )\\ &=\displaystyle \frac{-2\sin x\cos x}{2\left ( \cos ^{2}x \right )^{.^{\frac{1}{2}}}}\\ &=-\displaystyle \frac{\sin 2x}{2\sqrt{\cos ^{2}x}}\\ &=-\displaystyle \frac{\sin 2x}{2\left | \cos x \right |} \end{aligned} \end{array}$

CONTOH SOAL 5 TURUNAN PERTAMA FUNGSI TRIGONOMETRI

 

$\begin{array}{ll}\\ 21.&\textrm{Jika}\: \: h(x)=4x^{3}+\sin x+\cos x\\ &\textrm{maka}\: \: h'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&12x^{2}+\cos x-\sin x\\ \textrm{b}.&12x^{2}-\cos x+\sin x\\ \textrm{c}.&4x^{3}-\cos x-\sin x\\ \textrm{d}.&4x^{3}-\sin x-\cos x\\ \textrm{e}.&12x^{3}+\cos x+\sin x \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}h(x)&=4x^{3}+\sin x+\cos x\\ \textrm{guna}&\textrm{kan formula}:\: y=a.u^{n}\Rightarrow y'=n.a.u^{n-1}.u'\\ \textrm{pada}&\: \textrm{fungsi aljabarnya, yaitu}:\color{black}y=4x^{3}\Rightarrow y'=12x^{2}\\ \textrm{seda}&\textrm{ngkan fungsi transendennya mengikuti}\\ \textrm{turu}&\textrm{nan fungsi trigonometri biasa. Sehingga}\\ f'(x)&=12x^{2}+\cos x-\sin x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 22.&\textrm{Jika}\: \: p(x)=-\cos ^{4}x,\: \: \textrm{maka nilai}\\ &\textrm{maka}\: \: p'\left ( \displaystyle \frac{\pi }{3} \right )=....\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&\sqrt{3}\\ \textrm{c}.&\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{d}.&\displaystyle \frac{1}{4}\sqrt{3}\\ \textrm{e}.&1 \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}p(x)&=-\cos ^{4}x\\ p'(x)&=-4\cos ^{3}x.(-\sin x)=\color{black}4\cos ^{3}x\sin x\\ p'\left ( \displaystyle \frac{\pi }{3} \right )&=4\cos ^{3}\left ( \displaystyle \frac{\pi }{3} \right ).\sin \left ( \displaystyle \frac{\pi }{3} \right )\\ &=4\cos ^{3}60^{\circ}\times \sin 60^{\circ}\\ &=4\left ( \displaystyle \frac{1}{2} \right )^{3}\times \left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \frac{4}{16}\sqrt{3}\\ &=\displaystyle \frac{1}{4}\sqrt{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 23.&\textrm{Turunan pertama}\: \: q(x)=\sin ^{2}x+\cos ^{2}x\\ &\textrm{adalah}\: \: q'(x)=....\\ &\begin{array}{llll} \textrm{a}.&\cos ^{2}x-\sin ^{2}x\\ \textrm{b}.&2\cos ^{2}x-2\sin ^{2}x\\ \textrm{c}.&\cos x-\sin x\\ \textrm{d}.&2\cos x-2\sin x\\ \textrm{e}.&0 \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll}&\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}q(x)&=\sin ^{2}x+\cos ^{2}x\\ \textrm{guna}&\textrm{kan formula identitas}:\: \sin ^{2}x+\cos ^{2}x=1\\ \textrm{Sehi}&\textrm{ngga soal di atas dapat dituliskan menjadi}\\ q(x)&=1,\: \: \textrm{maka}\\ q'(x)&=0\\ \textrm{inga}&\textrm{t bahwa}\: \: y=a\Rightarrow \displaystyle \frac{dy}{dx}=0 \end{aligned}\end{array}$

$\begin{array}{ll}\\ 24.&\textrm{Nilai dari}\: \: \underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sin \left (\displaystyle \frac{\pi }{3}+h \right )-\sin \displaystyle \frac{\pi }{3}}{h}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{b}.&-\displaystyle \frac{1}{2}\\ \textrm{c}.&0\\ \textrm{d}.&\displaystyle \frac{1}{2}\\ \textrm{e}.&\displaystyle \frac{1}{2}\sqrt{3} \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Dari}&\: \textrm{soal diketahui}:\: \\ f(x)&=\sin \displaystyle \frac{\pi }{3}\\ \textrm{Nila}&\textrm{i dari}\: \: \underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sin \left (\displaystyle \frac{\pi }{3}+h \right )-\sin \displaystyle \frac{\pi }{3}}{h}\\ \textrm{arti}&\textrm{nya bermakna, berapkah}\: \: f'\left ( x \right )?\\ \textrm{maka}&\\ f'\left ( x \right )&=0 \end{aligned}\end{array}$

$\begin{array}{ll}\\ 25.&\textrm{Jika}\: \: f(x)=8x-\sin ^{3}x,\\ &\textrm{maka nilai}\: \: \underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\textrm{adalah}....\\ &\begin{array}{llll} \textrm{a}.&4x^{2}-3\cos^{2}x \\ \textrm{b}.&8x-3\sin ^{2}x\cos x\\ \textrm{c}.&8-3\sin ^{2}x\cos x\\ \textrm{d}.&8+\sin ^{2}x\cos x\\ \textrm{e}.&3\sin ^{2}x\cos x \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui dari soal}\: f(x)=8x-\sin ^{3}x\\ \textrm{maka}&\: \textrm{nilai dari}\: \: \underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{f(x+h)-f(x)}{h}=f'(x)\\ f'(x)&=8-3\sin ^{2}x\cos x \end{aligned}\end{array}$

CONTOH SOAL 4 TURUNAN PERTAMA FUNGSI TRIGONOMETRI

 

$\begin{array}{ll}\\ 16.&\textrm{Jika}\: \: W=\sin 2t\: ,\: \textrm{maka}\: \: \displaystyle \frac{dW}{dt}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \cos 2t&\textrm{d}.\quad 2t\cos 2t+\sin 2t\\ \textrm{b}.\quad 2\cos 2t\quad &\textrm{e}.\quad \sin 2t-t\cos 2t\\ \textrm{c}.\quad \sin 2t+t\cos 2t\quad \end{array} \end{array}$.

Jawaban: $\begin{array}{ll} &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &W=\sin 2t\\ &W=\sin u\, \quad \textrm{dengan}\: \: u=2t\\ &\displaystyle \frac{dW}{dt}=\displaystyle \frac{dW}{du}.\frac{du}{dt}\\ &\qquad=\cos u.2\\ &\qquad=2\cos u\\ &\qquad=2\cos 2t \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 17.&\textrm{Diketahui}\: \: f(x)=2\cos x-2020\\ &\textrm{Turunan pertama fungsi}\: \: f(x)\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&2\sin x\\ \textrm{b}.&-2\sin x\\ \textrm{c}.&-2\sin x-2020x\\ \textrm{d}.&2\sin ^{2}x\\ \textrm{e}.&2\cos x-2020x \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll}&\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}f(x)&=2\cos x-2020\\ f'(x)&=-2\sin x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Jika}\: \: f'(x)\: \: \textrm{adalah turunan pertama dari}\\ &\textrm{fungsi}\: \: f(x)=\sin ^{7}x\: ,\: \textrm{maka}\: \: f'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&7\cos^{6} x\\ \textrm{b}.&7\cos^{7} x\\ \textrm{c}.&7\sin^{6} x\cos x\\ \textrm{d}.&7\cos ^{6}x\sin x\\ \textrm{e}.&7\cos ^{6}x\sin ^{6}x \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll}&\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}f(x)&=\sin ^{7}x\\ \textrm{guna}&\textrm{kan formula}:\: y=a.u^{n}\Rightarrow y'=n.a.u^{n-1}.u'\\ f'(x)&=7\sin ^{6}x\left ( \cos x \right )=7\sin ^{6}x\cos x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 19.&\textrm{Turunan pertama fungsi}\: \: g(x)=-5\sin ^{3}x\\ &\textrm{adalah}\: \: g'(x)=....\\ &\begin{array}{llll} \textrm{a}.&-5\sin ^{2}x\cos x\\ \textrm{b}.&-5\sin ^{2}\cos ^{2}x\\ \textrm{c}.&-15\sin ^{2}x\cos x\\ \textrm{d}.&-15\cos ^{3}x\\ \textrm{e}.&-15\sin ^{4}x \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}g(x)&=-5\sin ^{3}x\\ \textrm{guna}&\textrm{kan formula}:\: y=a.u^{n}\Rightarrow y'=n.a.u^{n-1}.u'\\ g'(x)&=-5\left ( 3\sin ^{2}x \right )(\cos x)=-15\sin ^{2}x\cos x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Jika}\: \: \: f(x)=\displaystyle \frac{1+\sin x}{\cos x} \: ,\: \textrm{maka}\: \: {f}\, '\left ( \displaystyle \frac{1}{6}\pi \right )=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2} &&\textrm{d}.\quad 2 \\ \textrm{b}.\quad \displaystyle -\frac{1}{2} \quad &\textrm{c}.\quad \displaystyle \frac{3}{4} \quad &\textrm{e}.\quad -2 \end{array}\\\end{array}$.

Jawaban: $\begin{array}{ll}&\textbf{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle \frac{1+\sin x}{\cos x}\\ {f}\, '(x)&=\displaystyle \frac{\cos x.\cos x-(1+\sin x).-\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{\cos ^{2}x+\sin x +\sin ^{2}x}{\cos ^{2}x}\\ &\qquad \textrm{ingat bahwa}\: \: \sin ^{2}x+\cos ^{2}x=1\\ &=\displaystyle \frac{1+\sin x }{\cos ^{2}x}\\ {f}\, '\left ( \displaystyle \frac{1}{6}\pi \right )&=\displaystyle \frac{1+\sin \left ( \displaystyle \frac{1}{6}\pi \right )}{\cos ^{2}\left ( \displaystyle \frac{1}{6}\pi \right )}\\&=\displaystyle \frac{1+\displaystyle \frac{1}{2} }{\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )^{2}}\\ &=\displaystyle \frac{\displaystyle \frac{3}{2}}{\displaystyle \frac{3}{4}}\\ &=\displaystyle \frac{3}{2}\times \frac{4}{3}\\ &=2 \end{aligned} \end{array}$.

VISUALISASI GARIS SINGGUNG KURVA DENGAN GEOGEBRA MATERI TURUNAN FUNGSI

Petunjuk Simulasi Garis Singgung (GeoGebra):

1. Geser nilai slider a untuk berpindah posisi titik $A(a, f(a))$ pada kurva $f(x) = x^2$.

2. Perhatikan perubahan persamaan garis singgung $g$ serta gradiennya ($m = 2a$) di sepanjang kurva.

3. Saat titik $A$ berada di koordinat $(0,0)$, kemiringan garis bernilai $0$ ($f'(0) = 0$), sehingga terbentuk garis horizontal $y = 0$.

Mengenal GeoGebra: Dynamic Mathematics Software untuk Pembelajaran Interaktif

1. Apa itu GeoGebra?

GeoGebra adalah perangkat lunak matematika dinamis (dynamic mathematics software) yang menggabungkan geometri, aljabar, lembar kerja (spreadsheet), grafik, statistik, dan kalkulus dalam satu paket yang mudah digunakan. Nama GeoGebra sendiri merupakan gabungan dari kata Geometri dan Algebra.

GeoGebra dirancang khusus untuk menghubungkan representasi visual (geometris) dan representasi simbolis (aljabar) secara simultan. Ketika pengguna mengubah objek geometri di layar, nilai atau persamaan aljabarnya akan berubah secara otomatis, dan sebaliknya.

2. Kapan GeoGebra Mulai Dikembangkan dan Digunakan?

GeoGebra mulai dikembangkan pada tahun 2001 oleh Markus Hohenwarter sebagai proyek tesis magister dan disertasi doktornya di University of Salzburg, Austria.

Setelah dirilis sebagai perangkat lunak sumber terbuka (open-source), GeoGebra dengan cepat berkembang menjadi gerakan internasional. Pada tahun 2006, dikembangkan komunitas global serta *International GeoGebra Institute* (IGI) untuk mendukung pelatihan guru dan riset pendidikan. Saat ini, GeoGebra digunakan oleh puluhan juta siswa, mahasiswa, dan pendidik di lebih dari 190 negara.

3. Seberapa Penting GeoGebra untuk Dunia Pendidikan?

GeoGebra memegang peran yang sangat transformatif dalam dunia pendidikan matematika modern, antara lain:

• Koneksi Antar Konsep (Geometri & Aljabar): Siswa sering kali kesulitan menghubungkan rumus aljabar dengan bentuk geometrisnya. GeoGebra membatu siswa memvisualisasikan bagaimana persamaan seperti $f(x) = x^2$ berhubungan langsung dengan grafik parabola dan garis singgungnya.
• Eksplorasi dan Penemuan Mandiri (Guided Discovery): Melalui fitur slider dan konstruksi dinamis, GeoGebra mengubah siswa dari pembelajar pasif menjadi peneliti aktif yang bisa menguji hipotesis matematika secara mandiri.
• Gratis dan Lintas Platform: GeoGebra dapat diakses secara gratis di berbagai perangkat—baik melalui web browser, aplikasi desktop (Windows/Mac/Linux), maupun aplikasi seluler (Android/iOS).
• Ekosistem Bahan Ajar Berbagi (GeoGebra Materials): Komunitas GeoGebra menyediakan jutaan applet dan lembar kerja interaktif siap pakai yang dibuat oleh guru dari seluruh dunia, memudahkan pendidik dalam menyiapkan bahan ajar digital.

Ringkasan: GeoGebra bukan hanya sekadar alat bantu gambar grafik, melainkan lingkungan laboratorium matematika digital yang membantu siswa membangun intuisi konseptual secara visual dan matematis.

Daftar Pustaka / Referensi

• GeoGebra. (n.d.). About GeoGebra: Dynamic Mathematics for Everyone. Diterima dari https://www.geogebra.org/about
• Hohenwarter, M., & Preiner, J. (2007). Dynamic mathematics with GeoGebra. The Journal of Online Mathematics and its Applications, 7, 1-14.
• Hohenwarter, M., Hohenwarter, J., Kreis, Y., & Lavicza, Z. (2008). Teaching and learning calculus with GeoGebra. The Public Knowledge Journal, 1(1), 1-9.

Panduan Singkat Penggunaan GeoGebra untuk Pemula

Berikut adalah langkah-langkah dasar bagi siswa maupun pengajar untuk mengeksplorasi grafik dan konsep matematika secara interaktif menggunakan GeoGebra Graphing Calculator:

1. Memasukkan Persamaan Fungsi

Pada panel input di sebelah kiri, ketikkan fungsi matematika yang ingin dibuat. Contohnya: ketik f(x) = x^2 atau f(x) = sin(x), lalu tekan Enter. Kurva grafik akan langsung tergambar secara otomatis di lembar kerja.

2. Membuat Slider Interaktif

Slider digunakan untuk menggerakkan variabel secara dinamis. Ketik nama variabel di kolom input (contoh: a = 1 lalu tekan Enter). GeoGebra akan membuatkan tombol slider yang bisa digeser untuk mengubah nilai a secara langsung.

3. Menentukan Titik Singgung & Garis Singgung

Untuk menempatkan titik pada kurva, ketik A = (a, f(a)). Selanjutnya, untuk menggambar garis singgung yang menyinggung fungsi di titik tersebut, ketik perintah g = Tangent(A, f).

4. Mengatur Tampilan dan Navigasi Lembar Kerja

Gunakan fitur seret (click & drag) menggunakan mouse atau jari untuk menggeser area grafik. Anda juga dapat memperbesar/memperkecil tampilan menggunakan tombol roda mouse atau ikon + dan - di sudut kanan bawah.

💡 Tips Eksplorasi Siswa: Cobalah menggeser slider a dari nilai negatif ke positif, lalu perhatikan bagaimana persamaan garis singgung g dan kemiringannya (gradien) berubah secara otomatis sesuai turunan fungsinya.