Levi Rizki Saputra Notes

Turunan Fungsi Trigonometri

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# Turunan Sinus dan Cosinus

# Menggunakan Garis Singgung

Misalkan kita menggambar fungsi sin⁡(x)\sin(x) untuk menganalisis garis singgung fungsi pada berbagai titik terutama sudut istimewa.

Garis Singgung Fungsi Sinus

Maka derivasi dari sin⁡(x)\sin(x) pada berbagai titik berdasarkan kemiringan garis singgung.

xx d(sin⁡(x))/dxd(\sin(x))/dx
00 11
π/2\pi/2 00
π\pi −1-1
3π/23\pi/2 00
2π2\pi 11
…\dots …\ldots

Kemudian kita bisa menggambar y=d(sin⁡(x))/dxy = d(\sin(x))/dx dan mencoba menghubungkan titik-titik yang kita tahu untuk menebak fungsi adap d(sin⁡(x))/dxd(\sin(x))/dx.

Grafik Fungsi Turunan Sinus

Bisa ditebak secara intuitif berdasarkan grafik y=d(sin⁡(x))/dxy=d(\sin(x))/dx dan tabel nilai derivatif bahwa d(sin⁡(x))dx=cos⁡(x)\dfrac{d(\sin(x))}{dx}=\cos(x)

Metode ini juga dapat digunakan untuk mencari derivasi cos⁡(x)\cos(x).

Garis Singgung pada Fungsi Cosinus

Maka derivasi dari cos⁡(x)\cos(x) dari berbagai titik berdasarkan kemiringan garis singgung adalah

xx d(cos⁡(x))/dxd(\cos(x))/dx
00 00
π/2\pi/2 −1-1
π\pi 00
3π/23\pi/2 11
2π2\pi 00
…\dots …\ldots

Grafik Fungsi Turunan Cosinus

Dari data di atas kita bisa menggambar y=d(cos⁡(x))/dxy=d(\cos(x))/dx dan menghubungkan titik-titik pada grafik untuk menebak fungsi dari d(cos⁡(x))/dxd(\cos(x))/dx.

Bisa dilihat fungsi tersebut seperti fungsi sin yang dibalik nilai yy-nya (refleksi terhadap sumbu x). Bisa ditebak d(cos⁡(x))/dx=−sin⁡(x)d(\cos(x))/dx=-\sin(x).

# Turunan Sinus

# Menggunakan Limit Cara 1

d(sin⁡(x))dx=lim⁡h→0sin⁡(x+h)−sin⁡(x)h\frac{d(\sin(x))}{dx} =\lim_{h\to0}\frac{\sin(x+h)-\sin(x)}{h}

Kita bisa menggunakan sifat sin⁡(A−B)=2cos⁡(A+B2)sin⁡(A−B2)\sin(A-B)=2\cos\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)

d(sin⁡(x))dx=lim⁡h→0sin⁡(x+h)−sin⁡(x)h=lim⁡h→02cos⁡(x+h+x2)sin⁡(x+h−x2)h=lim⁡h→02cos⁡(2x+h2)sin⁡(h2)h=lim⁡h→02cos⁡(x+h2)sin⁡(h2)h=lim⁡h→0cos⁡(x+h2)sin⁡(h2)(h2)=lim⁡h→0cos⁡(x+h2)sin⁡(h2)(h2)=lim⁡h→0(cos⁡(x+h2))lim⁡h→0(sin⁡(h2)(h2))\begin{align*} \frac{d(\sin(x))}{dx} & =\lim_{h\to0}\frac{\sin(x+h)-\sin(x)}{h}\\ & =\lim_{h\to0}\frac{2\cos\left(\dfrac{x+h+x}{2}\right)\sin\left(\dfrac{x+h-x}{2}\right)}{h}\\ & =\lim_{h\to0}\frac{2\cos\left(\dfrac{2x+h}{2}\right)\sin\left(\dfrac{h}{2}\right)}{h}\\ & =\lim_{h\to0}\frac{2\cos\left(x+\dfrac{h}{2}\right)\sin\left(\dfrac{h}{2}\right)}{h}\\ & =\lim_{h\to0}\frac{\cos\left(x+\dfrac{h}{2}\right)\sin\left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\\ & =\lim_{h\to0}\cos\left(x+\dfrac{h}{2}\right)\frac{\sin\left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\\ & =\lim_{h\to0}\left(\cos\left(x+\dfrac{h}{2}\right)\right)\lim_{h\to0}\left(\frac{\sin\left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right) \end{align*}

Misalkan θ=h2\theta = \dfrac{h}{2}. Saat h→0h \to 0, h2→0\dfrac{h}{2} \to 0, maka θ→0\theta \to 0. Jadi lim⁡h→0=lim⁡θ→0\lim_{h\to0} = \lim_{\theta \to 0}

d(sin⁡(x))dx=lim⁡h→0(cos⁡(x+h2))lim⁡h→0(sin⁡(h2)(h2))=lim⁡h→0(cos⁡(x+h2))lim⁡θ→0(sin⁡(θ)θ)\begin{align*} \frac{d(\sin(x))}{dx} & =\lim_{h\to0}\left(\cos\left(x+\dfrac{h}{2}\right)\right)\lim_{h\to0}\left(\frac{\sin\left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right)\\ & =\lim_{h\to0}\left(\cos\left(x+\dfrac{h}{2}\right)\right)\lim_{\theta\to0}\left(\frac{\sin\left(\theta\right)}{\theta}\right) \end{align*}

Ingat sifat lim⁡θ→0sin⁡(θ)θ=1\lim_{\theta\to0}\dfrac{\sin(\theta)}{\theta} = 1. Lihat Limit Trigomometri

d(sin⁡(x))dx=lim⁡h→0(cos⁡(x+h2))lim⁡θ→0(sin⁡(θ)θ)=lim⁡h→0(cos⁡(x+h2))1=cos⁡(x+02)=cos⁡(x)\begin{align*} \frac{d(\sin(x))}{dx} & =\lim_{h\to0}\left(\cos\left(x+\dfrac{h}{2}\right)\right)\lim_{\theta\to0}\left(\frac{\sin\left(\theta\right)}{\theta}\right)\\ & =\lim_{h\to0}\left(\cos\left(x+\dfrac{h}{2}\right)\right)1\\ & =\cos(x+\frac{0}{2})\\ & =\cos(x) \end{align*}

# Menggunakan Limit Cara 2

Kita menggunakan sifat penjumlahan trigonometri.

d(sin⁡(x))dx=lim⁡h→0sin⁡(x+h)−sin⁡(x)h=lim⁡h→0sin⁡(x)cos⁡(h)+cos⁡(x)sin⁡(h)−sin⁡(x)h=lim⁡h→0sin⁡(x)(cos⁡(h)−1)+cos⁡(x)sin⁡(h)h=lim⁡h→0(sin⁡(x)(cos⁡(h)−1)h+cos⁡(x)sin⁡(h)h)=lim⁡h→0(sin⁡(x)(cos⁡(h)−1)h)+lim⁡h→0(cos⁡(x)sin⁡(h)h)=sin⁡(x)lim⁡h→0(cos⁡(h)−1h)+cos⁡(x)lim⁡h→0(sin⁡(h)h)\begin{align*} \frac{d(\sin(x))}{dx} & =\lim_{h\to0}\frac{\sin(x+h)-\sin(x)}{h}\\ & =\lim_{h\to0}\frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h}\\ & =\lim_{h\to0}\frac{\sin(x)(\cos(h)-1)+\cos(x)\sin(h)}{h}\\ & =\lim_{h\to0}\left(\frac{\sin(x)(\cos(h)-1)}{h}+\frac{\cos(x)\sin(h)}{h}\right)\\ & =\lim_{h\to0}\left(\frac{\sin(x)(\cos(h)-1)}{h}\right)+\lim_{h\to0}\left(\frac{\cos(x)\sin(h)}{h}\right)\\ & =\sin(x)\lim_{h\to0}\left(\frac{\cos(h)-1}{h}\right)+\cos(x)\lim_{h\to0}\left(\frac{\sin(h)}{h}\right) \end{align*}

Ingat sifat lim⁡θ→0cos⁡(θ)−1θ=0\lim_{\theta\to0}\dfrac{\cos(\theta)-1}{\theta} =0. Lihat Limit Trigomometri

d(sin⁡(x))dx=sin⁡(x)lim⁡h→0(cos⁡(h)−1h)+cos⁡(x)lim⁡h→0(sin⁡(h)h)=sin⁡(x)×0+cos⁡(x)×1=cos⁡(x)\begin{align*} \frac{d(\sin(x))}{dx} & =\sin(x)\lim_{h\to0}\left(\frac{\cos(h)-1}{h}\right)+\cos(x)\lim_{h\to0}\left(\frac{\sin(h)}{h}\right)\\ & =\sin(x)\times0+\cos(x)\times1\\ & =\cos(x) \end{align*}

# Turunan Cosinus

# Menggunakan Sifat Trigonometri

d(cos⁡(x))dx=d(sin⁡(90°−x))dx=d(sin⁡(π2−x))dx=d(sin⁡(π2−x))d(π2−x)×(π2−x)dx=cos⁡(π2−x)×(0−1)=−cos⁡(π2−x)=−sin⁡(x)\begin{align*} \frac{d(\cos(x))}{dx} & =\frac{d(\sin(90\degree-x))}{dx}\\ & =\frac{d\left(\sin\left(\dfrac{\pi}{2}-x\right)\right)}{dx}\\ & =\frac{d\left(\sin\left(\dfrac{\pi}{2}-x\right)\right)}{d\left(\dfrac{\pi}{2}-x\right)}\times\frac{\left(\dfrac{\pi}{2}-x\right)}{dx}\\ & =\cos\left(\frac{\pi}{2}-x\right)\times(0-1)\\ & =-\cos\left(\frac{\pi}{2}-x\right)\\ & =-\sin(x) \end{align*}

# Menggunakan Limit

Kita menggunakan sifat penjumlahan cosinus

d(cos⁡(x))dx=lim⁡h→0cos⁡(x+h)−cos⁡(x)h=lim⁡h→0cos⁡(x)cos⁡(h)−sin⁡(x)sin⁡(h)−cos⁡(x)h=lim⁡h→0cos⁡(x)(cos⁡(h)−1)−sin⁡(x)sin⁡(h)h=lim⁡h→0(cos⁡(x)(cos⁡(h)−1)h−sin⁡(x)sin⁡(h)h)=lim⁡h→0(cos⁡(x)(cos⁡(h)−1)h)−lim⁡h→0(sin⁡(x)sin⁡(h)h)=cos⁡(x)lim⁡h→0((cos⁡(h)−1)h)−sin⁡(x)lim⁡h→0(sin⁡(h)h)=cos⁡(x)×0−sin⁡(x)×1=−sin⁡(x)\begin{align*} \frac{d(\cos(x))}{dx} & =\lim_{h\to0}\frac{\cos(x+h)-\cos(x)}{h}\\ & =\lim_{h\to0}\frac{\cos(x)\cos(h)-\sin(x)\sin(h)-\cos(x)}{h}\\ & =\lim_{h\to0}\frac{\cos(x)(\cos(h)-1)-\sin(x)\sin(h)}{h}\\ & =\lim_{h\to0}\left(\cos(x)\frac{(\cos(h)-1)}{h}-\sin(x)\frac{\sin(h)}{h}\right)\\ & =\lim_{h\to0}\left(\cos(x)\frac{(\cos(h)-1)}{h}\right)-\lim_{h\to0}\left(\sin(x)\frac{\sin(h)}{h}\right)\\ & =\cos(x)\lim_{h\to0}\left(\frac{(\cos(h)-1)}{h}\right)-\sin(x)\lim_{h\to0}\left(\frac{\sin(h)}{h}\right)\\ & =\cos(x)\times0-\sin(x)\times1\\ & =-\sin(x) \end{align*}

# Derivasi Tangen

d(tan(x))dx=ddx(sin⁡(x)cos⁡(x))=d(sin⁡(x))dxcos⁡(x)−d(cos⁡(x))dxsin⁡(x)cos⁡2(x)=cos⁡(x)cos⁡(x)−(−sin⁡(x))sin⁡(x)cos⁡2(x)=cos⁡2(x)+sin⁡2(x)cos⁡2(x)=1cos⁡2(x)=sec⁡2(x)\begin{align*} \frac{d(tan(x))}{dx} & =\frac{d}{dx}\left(\frac{\sin(x)}{\cos(x)}\right)\\ & =\dfrac{\frac{d(\sin(x))}{dx}\cos(x)-\frac{d(\cos(x))}{dx}\sin(x)}{\cos^{2}(x)}\\ & =\dfrac{\cos(x)\cos(x)-(-\sin(x))\sin(x)}{\cos^{2}(x)}\\ & =\dfrac{\cos^{2}(x)+\sin^{2}(x)}{\cos^{2}(x)}\\ & =\frac{1}{\cos^{2}(x)}\\ & =\sec^{2}(x) \end{align*}

# Derivasi Secan

d(sec⁡(x))dx=ddx(1cos⁡(x))=d(1)dxcos⁡(x)−d(cos⁡(x))dx×1cos⁡2(x)=0×cos⁡(x)−(−sin⁡(x))cos⁡2(x)=sin⁡(x)cos⁡2(x)=sin⁡(x)cos⁡(x)cos⁡(x)=tan⁡(x)sec⁡(x)\begin{align*} \frac{d(\sec(x))}{dx} & =\frac{d}{dx}\left(\frac{1}{\cos(x)}\right)\\ & =\dfrac{\frac{d(1)}{dx}\cos(x)-\frac{d(\cos(x))}{dx}\times1}{\cos^{2}(x)}\\ & =\dfrac{0\times\cos(x)-(-\sin(x))}{\cos^{2}(x)}\\ & =\dfrac{\sin(x)}{\cos^{2}(x)}\\ & =\frac{\sin(x)}{\cos(x)\cos(x)}\\ & =\tan(x)\sec(x) \end{align*}

# Derivasi Cosecan

d(csc⁡(x))dx=ddx(1sin⁡(x))=d(1)dxsin⁡(x)−d(sin⁡(x))dx×1sin⁡2(x)=0×sin⁡(x)−cos⁡(x)sin⁡2(x)=−cos⁡(x)sin⁡2(x)=−cos⁡(x)sin⁡(x)sin⁡(x)=−cot⁡(x)csc⁡(x)\begin{align*} \frac{d(\csc(x))}{dx} & =\frac{d}{dx}\left(\frac{1}{\sin(x)}\right)\\ & =\dfrac{\frac{d(1)}{dx}\sin(x)-\frac{d(\sin(x))}{dx}\times1}{\sin^{2}(x)}\\ & =\dfrac{0\times\sin(x)-\cos(x)}{\sin^{2}(x)}\\ & =\dfrac{-\cos(x)}{\sin^{2}(x)}\\ & =\frac{-\cos(x)}{\sin(x)\sin(x)}\\ & =-\cot(x)\csc(x) \end{align*}

# Derivasi Cotangen

d(tan⁡(x))dx=ddx(cos⁡(x)sin⁡(x))=d(cos⁡(x))dxsin⁡(x)−d(sin⁡(x))dxcos⁡(x)sin⁡2(x)=(−sin⁡(x))sin⁡(x)−cos⁡(x)cos⁡(x)sin⁡2(x)=−sin⁡2(x)−cos⁡2(x)sin⁡2(x)=−(sin⁡2(x)+cos⁡2(x))sin⁡2(x)=−1sin⁡2(x)=−csc⁡2(x)\begin{align*} \frac{d(\tan(x))}{dx} & =\frac{d}{dx}\left(\frac{\cos(x)}{\sin(x)}\right)\\ & =\dfrac{\frac{d(\cos(x))}{dx}\sin(x)-\frac{d(\sin(x))}{dx}\cos(x)}{\sin^{2}(x)}\\ & =\dfrac{(-\sin(x))\sin(x)-\cos(x)\cos(x)}{\sin^{2}(x)}\\ & =\dfrac{-\sin^{2}(x)-\cos^{2}(x)}{\sin^{2}(x)}\\ & =\dfrac{-(\sin^{2}(x)+\cos^{2}(x))}{\sin^{2}(x)}\\ & =\frac{-1}{\sin^{2}(x)}\\ & =-\csc^{2}(x) \end{align*}

# Kesimpulan

Derivasi untuk fungsi trigonometri

d(sin⁡(x))dx=cos⁡(x)d(cos⁡(x))dx=−sin⁡(x)d(tan⁡(x))dx=sec⁡2(x)d(sec⁡(x))dx=tan⁡(x)sec⁡(x)d(csc⁡(x))dx=−cot⁡(x)csc⁡(x)d(cot⁡(x))dx=−csc⁡2(x)\begin{align*} \frac{d(\sin(x))}{dx} & =\cos(x)\\ \frac{d(\cos(x))}{dx} & =-\sin(x)\\ \frac{d(\tan(x))}{dx} & =\sec^{2}(x)\\ \frac{d(\sec(x))}{dx} & =\tan(x)\sec(x)\\ \frac{d(\csc(x))}{dx} & =-\cot(x)\csc(x)\\ \frac{d(\cot(x))}{dx} & =-\csc^{2}(x) \end{align*}