Materi, Soal, dan Pembahasan – Turunan Fungsi Implisit

Diferensialkan setiap suku terhadap $x.$
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x} (ax^2)+\dfrac{\text{d}}{\text{d}x} (by^2) & = \dfrac{\text{d}}{\text{d}x} (1) \\ 2ax+\left(0 + 2by~\dfrac{\text{d}y}{\text{d}x}\right) & = 0 \\ 2by~\dfrac{\text{d}y}{\text{d}x} & = -2ax \\ \dfrac{\text{d}y}{\text{d}x} & = \dfrac{-2ax}{2by} \\ \dfrac{\text{d}y}{\text{d}x} & = -\dfrac{ax}{by} \end{aligned}$$Jadi, diperoleh $\boxed{\dfrac{\text{d}y}{\text{d}x} = -\dfrac{ax}{by}}.$

Diferensialkan setiap suku terhadap $x.$
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x} (\sqrt{x})+\dfrac{\text{d}}{\text{d}x} (\sqrt{y}) & = \dfrac{\text{d}}{\text{d}x} (a) \\ \dfrac{1}{2\sqrt{x}} +\left(0 + \dfrac{1}{2\sqrt{y}}~\dfrac{\text{d}y}{\text{d}x}\right) & = 0 \\ \dfrac{1}{2\sqrt{y}}~\dfrac{\text{d}y}{\text{d}x} & = -\dfrac{1}{2\sqrt{x}} \\ \dfrac{\text{d}y}{\text{d}x} & = -\dfrac{1}{2\sqrt{x}} \cdot 2\sqrt{y} \\ \dfrac{\text{d}y}{\text{d}x} & = -\sqrt{\dfrac{y}{x}} \end{aligned}$$Jadi, diperoleh $\boxed{\dfrac{\text{d}y}{\text{d}x} = -\sqrt{\dfrac{y}{x}}}.$

Diferensialkan setiap suku terhadap $x.$
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x} (\sin x)+\dfrac{\text{d}}{\text{d}x} (\sin y) & = \dfrac{\text{d}}{\text{d}x} (\pi) \\ \cos x +\left(0 + \cos y~\dfrac{\text{d}y}{\text{d}x}\right) & = 0 \\ \cos y~\dfrac{\text{d}y}{\text{d}x} & = -\cos x \\ \dfrac{\text{d}y}{\text{d}x} & = -\dfrac{\cos x}{\cos y} \end{aligned}$$Jadi, diperoleh $\boxed{\dfrac{\text{d}y}{\text{d}x} = -\dfrac{\cos x}{\cos y}}.$

Perhatikan bahwa $3xy-4 = x$ ekuivalen dengan $y = \dfrac{x+4}{3x}$. Sekarang, $y$ dinyatakan sebagai fungsi eksplisit dari $x$ dan akan kita cari turunannya menggunakan aturan hasil bagi.
Misalkan
$$\begin{aligned} u & = x + 4 \Rightarrow u’  = 1. \\ v & = 3x \Rightarrow v’ = 3. \end{aligned}$$Kita peroleh
$$\begin{aligned} \dfrac{\text{d}y}{\text{d}x} & = \dfrac{u’v-uv’}{v^2} \\ & = \dfrac{1(3x)-(x+4)(3)}{(3x)^2} \\ & = \dfrac{3x-3x-12}{9x^2} \\ & = \dfrac{-12}{9x^2} = \color{blue}{-\dfrac{4}{3x^2}}. \end{aligned}$$Sekarang kita akan menurunkan secara implisit fungsi $3xy-4 = x.$
Diferensialkan setiap suku terhadap $x.$
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x} (3xy)-\dfrac{\text{d}}{\text{d}x}(4) & = \dfrac{\text{d}}{\text{d}x} (x) \\ \left(3y + 3x~\dfrac{\text{d}y}{\text{d}x}\right)-0 & = 1 \\ 3x~\dfrac{\text{d}y}{\text{d}x} & = 1-3y \\ \dfrac{\text{d}y}{\text{d}x} & = \dfrac{1-3\color{red}{y}}{3x} \end{aligned}$$Sekarang kita peroleh turunan implisitnya, tetapi perhatikan bahwa ruas kanannya masih memuat variabel $y.$ Oleh karena itu, kita perlu lakukan substitusi $y = \dfrac{x+4}{3x}.$
$$\begin{aligned} \dfrac{\text{d}y}{\text{d}x} & = \dfrac{1-3\left(\dfrac{x+4}{3x}\right)}{3x} \\ & = \dfrac{\dfrac{3x}{3x}-\dfrac{3x+12}{3x}}{3x} \\ & = \dfrac{-12}{9x^2} = \color{blue}{-\dfrac{4}{3x^2}} \end{aligned}$$Jadi, terbukti bahwa fungsi $3xy-4 = x$ memiliki turunan yang sama terhadap $x$ bila dalam bentuk fungsi eksplisit maupun fungsi implisit.

Perhatikan bahwa fungsi di atas dapat ditulis sebagai
$$\begin{aligned} (x^2-6x+9)y & = 4 \\ (x-3)^2y & = 4 \\ y & = \dfrac{4}{(x-3)^2}. \end{aligned}$$Berikutnya, $y$ dinyatakan sebagai fungsi eksplisit dari $x$ dan akan kita cari turunannya menggunakan aturan basil bagi.
Misalkan
$$\begin{aligned} u & = 4 \Rightarrow u’ = 0. \\ v & = (x-3)^2 \Rightarrow v’ = 2(x-3) = 2x-6. \end{aligned}$$Kita peroleh
$$\begin{aligned} \dfrac{\text{d}y}{\text{d}x} & = \dfrac{u’v-uv’}{v^2} \\ & = \dfrac{0(x-3)^2-4(2x-6)}{\left((x-3)^2)\right)^2} \\ & = \color{blue}{\dfrac{-8x+24}{(x-3)^4}}. \end{aligned}$$Kita akan menurunkan secara implisit fungsi $x^2y-6xy+9y=4.$
Diferensialkan setiap suku terhadap $x.$
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x} (x^2y)-\dfrac{\text{d}}{\text{d}x}(6xy) + \dfrac{\text{d}}{\text{d}x} (9y) & = \dfrac{\text{d}}{\text{d}x} (4) \\ \left(2xy + x^2~\dfrac{\text{d}y}{\text{d}x}\right)-6\left(y+x~\dfrac{\text{d}y}{\text{d}x}\right) + 9~\dfrac{\text{d}y}{\text{d}x} & = 0 \\ (x^2-6x+9)~\dfrac{\text{d}y}{\text{d}x} & = -2xy+6y \\ (x-3)^2\dfrac{\text{d}y}{\text{d}x} & = (-2x+6)y \\ \dfrac{\text{d}y}{\text{d}x} & = \dfrac{(-2x+6)y}{(x-3)^2} \end{aligned}$$Kita peroleh turunan implisitnya, tetapi perhatikan bahwa ruas kanannya masih memuat variabel $y$. Oleh karena itu, kita perlu lakukan substitusi $y = \dfrac{4}{(x-3)^2}.$
$$\begin{aligned} \dfrac{\text{d}y}{\text{d}x} & = \dfrac{(-2x+6)\left(\dfrac{4}{(x-3)^2}\right)}{(x-3)^2} \\ & = \color{blue}{\dfrac{-8x+24}{(x-3)^4}} \end{aligned}$$Jadi, terbukti bahwa fungsi $x^2y-6xy+9y = 4$ memiliki turunan yang sama terhadap $x$ jika dalam bentuk fungsi eksplisit maupun fungsi implisit.

Misalkan $f(x, y) = (x+2y)^8$ sehingga berdasarkan aturan rantai dan aturan turunan fungsi implisit, diperoleh
$$\begin{aligned} f'(x, y) & = 8(x+2y)^7\left(x~\dfrac{\text{d}}{\text{d}x} + 2y~\dfrac{\text{d}}{\text{d}y}~\dfrac{\text{d}y}{\text{d}x}\right) \\ & = 8(x+2y)^7\left(1+2~\dfrac{\text{d}y}{\text{d}x}\right) \\ & = 8(x+2y)^7 + 16(x+2y)^7~\dfrac{\text{d}y}{\text{d}x} \end{aligned}$$Jadi, turunan pertama fungsi implisit itu adalah $\boxed{8(x+2y)^7 + 16(x+2y)^7~\dfrac{\text{d}y}{\text{d}x}}.$

Fungsi di atas dapat ditulis dalam bentuk implisit sebagai berikut.
$$\begin{aligned} y & = \dfrac{x}{x^2+1} \\ y(x^2+1) & = x \\ x^2y + y-x & = 0 \end{aligned}$$Masing-masing suku didiferensialkan terhadap variabel $x$ sehingga kita peroleh
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x}(x^2y) +\dfrac{\text{d}}{\text{d}x} (y)-\dfrac{\text{d}}{\text{d}x}(x) & = 0 \\ \left(x^2~\dfrac{\text{d}y}{\text{d}x} + 2xy\right) + \dfrac{\text{d}y}{\text{d}x}-1 & = 0 \\ (x^2+1)~\dfrac{\text{d}y}{\text{d}x} & =1-2xy \\ \dfrac{\text{d}y}{\text{d}x} & = \dfrac{1-2xy}{x^2+1}. \end{aligned}$$Jadi, turunan pertamanya adalah $\boxed{\dfrac{\text{d}y}{\text{d}x} = \dfrac{1-2xy}{x^2+1}}.$

Masing-masing suku didiferensialkan terhadap variabel $x$. Gunakan aturan hasil kali untuk menghitung turunan suku pertama dan aturan rantai untuk suku kedua.
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x} (xy) + \dfrac{\text{d}}{\text{d}x} (x+y+1)^3 & = \dfrac{\text{d}}{\text{d}x} (0) \\ \left(x~\dfrac{\text{d}y}{\text{d}x} + y\right) + 3(x+y+1)^2~\dfrac{\text{d}}{\text{d}x}(x+y+1) & = 0 \\ x~\dfrac{\text{d}y}{\text{d}x} + y + 3(x+y+1)^2\left(1+\dfrac{\text{d}y}{\text{d}x}\right) & = 0 \\ x~\dfrac{\text{d}y}{\text{d}x} + y + 3(x+y+1)^2 + 3(x+y+1)^2~\dfrac{\text{d}y}{\text{d}x} & = 0 \\ (x + 3(x+y+1)^2)~\dfrac{\text{d}y}{\text{d}x} & = -\left(y + 3(x+y+1)^2\right) \\ \dfrac{\text{d}y}{\text{d}x} & = -\dfrac{y + 3(x+y+1)^2}{x + 3(x+y+1)^2} \end{aligned}$$Jadi, hasil dari $\boxed{\dfrac{\text{d}y}{\text{d}x} = -\dfrac{y + 3(x+y+1)^2}{x + 3(x+y+1)^2}}.$

Diketahui $$x^2+y^2-5x+8y+\color{red}{2xy^2} = 19.$$Kita akan mencari turunan untuk suku $2xy^2$ terhadap $x$ terlebih dahulu agar penulisan nantinya tidak terlalu panjang.
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x} (2xy^2) & = 2 \dfrac{\text{d}}{\text{d}x} (xy^2) \\ & = 2\left(\dfrac{\text{d}}{\text{d}x} (x) \cdot y^2 + x \cdot \dfrac{\text{d}}{\text{d}x} (y^2)\right) && (\text{Aturan hasil kali}) \\ & = 2\left(1 \cdot y^2 + x \cdot 2y~\dfrac{\text{d}y}{\text{d}x}\right) \\ & = 2\left(y^2 + 2xy~\dfrac{\text{d}y}{\text{d}x}\right) \end{aligned}$$Dengan demikian, turunan pertama fungsi implisit tersebut secara keseluruhan adalah
$$\begin{aligned} x^2~\dfrac{\text{d}}{\text{d}x} + y^2~\dfrac{\text{d}}{\text{d}y}~\dfrac{\text{d}y}{\text{d}x}-5x~\dfrac{\text{d}}{\text{d}x} + 8y~\dfrac{\text{d}}{\text{d}y}~\dfrac{\text{d}y}{\text{d}x} + 2y^2 + 4xy~\dfrac{\text{d}y}{\text{d}x} & = 19~\dfrac{\text{d}}{\text{d}x} \\ 2x + 2y~\dfrac{\text{d}y}{\text{d}x}-5 + 8~\dfrac{\text{d}y}{\text{d}x} + 2y^2 + 4xy~\dfrac{\text{d}y}{\text{d}x} & = 0 \\ (2y + 8 + 4xy)~\dfrac{\text{d}y}{\text{d}x} + 2x-5+2y^2 & = 0 \\ (2y+8+4xy)~\dfrac{\text{d}y}{\text{d}x} & = 5-2x-2y^2 \\ \dfrac{\text{d}y}{\text{d}x} & = \dfrac{5-2x-2y^2}{2y+8+4xy} \end{aligned}$$Jadi, turunan pertama dalam bentuk $\dfrac{\text{d}y}{\text{d}x}$ dari fungsi implisit $$x^2+y^2-5x+8y+2xy^2 = 19$$ adalah $\boxed{\dfrac{\text{d}y}{\text{d}x} = \dfrac{5-2x-2y^2}{2y+8+4xy}}.$ 

Diketahui $\sin (xy) + xy^2 + x^2y = 1.$
Dengan menggunakan aturan turunan fungsi implisit terhadap variabel $x$, kita peroleh
$$\begin{aligned} \sin (xy) + xy^2 + x^2y & = 1 \\ \left(\sin (xy)~\dfrac{\text{d}}{\text{d}x} + \sin xy~\dfrac{\text{d}}{\text{d}y}~\dfrac{\text{d}y}{\text{d}x}\right)+\left(xy^2~\dfrac{\text{d}}{\text{d}x}+xy^2~\dfrac{\text{d}}{\text{d}y} ~\dfrac{\text{d}y}{\text{d}x}\right)+\left(x^2y~\dfrac{\text{d}}{\text{d}x} + x^2y~\dfrac{\text{d}}{\text{d}y}~\dfrac{\text{d}y}{\text{d}x}\right) & = 1~\dfrac{\text{d}}{\text{d}x} \\ \left(y \cos (xy) + x \cos xy~\dfrac{\text{d}y}{\text{d}x}\right)+\left(y^2+2xy~\dfrac{\text{d}y}{\text{d}x}\right) + \left(2xy + x^2~\dfrac{\text{d}y}{\text{d}x}\right) & = 0 \\ y(\cos (xy) + y + 2x) + (x \cos (xy) + 2xy + x^2)~\dfrac{\text{d}y}{\text{d}x} & = 0 \\ (x \cos (xy) + 2xy + x^2)~\dfrac{\text{d}y}{\text{d}x} & = -y(\cos (xy) + y + 2x) \\ \dfrac{\text{d}y}{\text{d}x} & = \dfrac{-y(\cos (xy)+y+2x)}{x \cos (xy) + 2xy + x^2}. \end{aligned}$$Jadi, turunan pertama dalam bentuk $\dfrac{\text{d}y}{\text{d}x}$ dari fungsi implisit $\sin (xy) + xy^2 + x^2y = 1$ adalah $\boxed{\dfrac{\text{d}y}{\text{d}x} = \dfrac{-y(\cos xy+y+2x)}{x \cos (xy) + 2xy + x^2}}.$

Fungsi $f$ dinyatakan dalam bentuk pecahan sehingga kita dapat menggunakan aturan hasil bagi untuk menentukan turunannya.
Misalkan
$$\begin{aligned} u & = y-x^2 \Rightarrow u’ = \dfrac{\text{d}y}{\text{d}x}-2x \\ v & = y^2-x \Rightarrow v’ = 2y~\dfrac{\text{d}y}{\text{d}x}-1. \end{aligned}$$Dengan demikian, kita peroleh
$$\begin{aligned} f'(x, y) & = \dfrac{u’v -uv’}{v^2} \\ & = \dfrac{\left(\dfrac{\text{d}y}{\text{d}x}-2x\right)(y^2-x)-(y-x^2)\left(2y~\dfrac{\text{d}y}{\text{d}x}-1\right)}{(y^2-x)^2} \\ & = \dfrac{(y^2-x-2y^2+2x^2y)~\dfrac{\text{d}y}{\text{d}x}-2xy^2+2x^2+y-x^2}{(y^2-x)^2} \\ & = \dfrac{(-x-y^2+2x^2y)~\dfrac{\text{d}y}{\text{d}x}-2xy^2+x^2+y}{(y^2-x)^2}. \end{aligned}$$Jadi, turunan pertama fungsi implisit $f(x, y) = \dfrac{y-x^2}{y^2-x}$ terhadap variabel $x$ adalah $$\boxed{f'(x, y) = \dfrac{(-x-y^2+2x^2y)~\dfrac{\text{d}y}{\text{d}x}-2xy^2+x^2+y}{(y^2-x)^2}}.$$

Diferensialkan setiap suku terhadap $x.$
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x} (x^3)-\dfrac{\text{d}}{\text{d}x} (xy) + \dfrac{\text{d}}{\text{d}x} (y^3) & = \dfrac{\text{d}}{\text{d}x} (1) \\ 3x^2-\left(x~\dfrac{\text{d}y}{\text{d}x} + y\right) + 3y^2~\dfrac{\text{d}y}{\text{d}x} & = 0 \\ (-x+3y^2)~\dfrac{\text{d}y}{\text{d}x} + 3x^2-y & = 0 \\ \dfrac{\text{d}y}{\text{d}x} & = \dfrac{y-3x^2}{3y^2-x} \end{aligned}$$Jadi, diperoleh $\boxed{\dfrac{\text{d}y}{\text{d}x} = \dfrac{y-3x^2}{3y^2-x}}.$

Diferensialkan setiap suku terhadap $x.$
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x} (3x+7)^5 & = \dfrac{\text{d}}{\text{d}x} (2y^3) \\ 5(3x+7)^4 \cdot \underbrace{3}_{\text{Turunan}~(3x+7)} & = 6y^2~\dfrac{\text{d}y}{\text{d}x} \\ \dfrac{\text{d}y}{\text{d}x} & = \dfrac{5(3x+7)^4}{2y^2} \end{aligned}$$Jadi, diperoleh $\boxed{\dfrac{\text{d}y}{\text{d}x} = \dfrac{5(3x+7)^4}{2y^2}}.$

Sederhanakan terlebih dahulu dengan cara menguraikan bentuk kubiknya (ruas kiri).
$$\begin{aligned} (x+y)^3 + (x-y)^3 & = x^4+y^4 \\ (x^3+3x^2y + 3xy^2 + y^3) + (x^3-3x^2y + 3xy^2-y^3) & = x^4+y^4 \\ 2x^3 + 6xy^2 & = x^4+y^4 \end{aligned}$$Diferensialkan setiap suku terhadap $x.$
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x} (2x^3) + \dfrac{\text{d}}{\text{d}x} (6xy^2) & = \dfrac{\text{d}}{\text{d}x} (x^4) + \dfrac{\text{d}}{\text{d}x} (y^4) \\ 6x^2 + 6\left(2xy~\dfrac{\text{d}y}{\text{d}x} + y^2\right) & = 4x^3 + 4y^3~\dfrac{\text{d}y}{\text{d}x} \\ 3x^2 + 3\left(2xy~\dfrac{\text{d}y}{\text{d}x} + y^2\right) & = 2x^3 + 2y^3~\dfrac{\text{d}y}{\text{d}x} && (\text{dibagi}~2) \\ (6xy-2y^3)~\dfrac{\text{d}y}{\text{d}x} & = 2x^3-3x^2-3y^2 \\ \dfrac{\text{d}y}{\text{d}x} & = \dfrac{2x^3-3x^2-3y^2}{6xy-2y^3} \end{aligned}$$Jadi, diperoleh $$\boxed{\dfrac{\text{d}y}{\text{d}x} = \dfrac{2x^3-3x^2-3y^2}{6xy-2y^3}}.$$

Pertama, kita dapat tuliskan persamaan di atas dalam bentuk
$$\begin{aligned} y^3(x+y) & = x-y \\ xy^3 + y^4 & = x-y. \end{aligned}$$Selanjutnya, diferensialkan terhadap variabel $x.$
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x} (xy^3) + \dfrac{\text{d}}{\text{d}x} (y^4) & = \dfrac{\text{d}}{\text{d}x} (x)-\dfrac{\text{d}}{\text{d}x} (y) \\ \left(3xy^2~\dfrac{\text{d}y}{\text{d}x} + y^3\right) + 4y^3~\dfrac{\text{d}y}{\text{d}x} & = 1-\dfrac{\text{d}y}{\text{d}x} \\ (3xy^2 + 4y^3 + 1)~\dfrac{\text{d}y}{\text{d}x} & = 1-y^3 \\ \dfrac{\text{d}y}{\text{d}x} & = \dfrac{1-y^3}{1+3xy^2+4y^3} \end{aligned}$$Jadi, diperoleh $\boxed{\dfrac{\text{d}y}{\text{d}x} = \dfrac{1-y^3}{1+3xy^2+4y^3}}.$ 

Diketahui
$$\begin{aligned} x^3 + 5 \ln (xy)-3xy^{-1} & = -4 \\ x^3 + 5 \ln x + 5 \ln y-3xy^{-1} & = -4. \end{aligned}$$Masing-masing suku didiferensialkan terhadap variabel $x$ sehingga kita peroleh
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x}~(x^3)+ \dfrac{\text{d}}{\text{d}x} \left(5 \ln x\right)+\dfrac{\text{d}}{\text{d}x} \left(5 \ln y\right)-\dfrac{\text{d}}{\text{d}x}(3xy^{-1}) & = \dfrac{\text{d}}{\text{d}x} (-4) \\ 3x^2 + \dfrac{5}{x} + \dfrac{5}{y}~\dfrac{\text{d}y}{\text{d}x}-3\left(y^{-1} + (-1)xy^{-2}~\dfrac{\text{d}y}{\text{d}x}\right) & = 0 \\ 3x^2 + 5x^{-1} + 5y^{-1}~\dfrac{\text{d}y}{\text{d}x}-3y^{-1}+3xy^{-2}~\dfrac{\text{d}y}{\text{d}x} & = 0 \\ \left(5y^{-1} + 3xy^{-2}\right)~\dfrac{\text{d}y}{\text{d}x} & = -3x^2-5x^{-1}+3y^{-1} \\ \dfrac{\text{d}y}{\text{d}x} & = \dfrac{-3x^2-5x^{-1}+3y^{-1}}{5y^{-1}+3xy^{-2}}. \end{aligned}$$Jadi, diperoleh $\boxed{\dfrac{\text{d}y}{\text{d}x} = \dfrac{-3x^2-5x^{-1}+3y^{-1}}{5y^{-1}+3xy^{-2}}}.$

Persamaan semula dapat disederhanakan dengan cara dibagi $y$ pada kedua ruasnya menjadi $$x^2y + 4x = 12$$dengan syarat $y \neq 0.$
Diferensialkan setiap suku terhadap variabel $x.$
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x} (x^2y) + \dfrac{\text{d}}{\text{d}x} (4x) & = \dfrac{\text{d}y}{\text{d}x} (12) \\ \left(2xy + x^2~\dfrac{\text{d}y}{\text{d}x}\right) + 4 & = 0 \\ x^2~\dfrac{\text{d}y}{\text{d}x} & = -2xy-4 \\ \dfrac{\text{d}y}{\text{d}x} & = \dfrac{-2xy-4}{x^2} \end{aligned}$$Gradien garis singgung diperoleh saat substitusi $x = 2$ dan $y = 1$ pada bentuk $\dfrac{\text{d}y}{\text{d}x}$ di atas, yaitu
$$\begin{aligned} m & = \left.\dfrac{\text{d}y}{\text{d}x}\right|_{(x,y) = (2,1)} \\ & = \dfrac{-2(2)(1)-4}{2^2} \\ & = \dfrac{-8}{4} = -2. \end{aligned}$$Persamaan garis singgung yang melalui titik $(x_1, y_1) = (2, 1)$ dan bergradien $m = -2$ dinyatakan oleh
$$\begin{aligned} y & = m(x-x_1)+y_1 \\ y & = -2(x-2) + 1 \\ y & = -2x + 5. \end{aligned}$$Jadi, persamaan garis singgung kurva $x^2y^2 + 4xy = 12y$ di titik $(2, 1)$ adalah $$\boxed{y = -2x + 5}.$$Berikut gambar grafik kurva dan garis singgungnya (menggunakan aplikasi GeoGebra).

Diferensialkan setiap suku persamaan tersebut terhadap variabel $x.$
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x} (x^3y) + \dfrac{\text{d}}{\text{d}x} (xy^3) & = \dfrac{\text{d}y}{\text{d}x} (10) \\ \left(3x^2y + x^3~\dfrac{\text{d}y}{\text{d}x}\right)+\left(y^3 + 3x^2y~\dfrac{\text{d}y}{\text{d}x}\right) & = 0 \\ (x^3+3xy^2)~\dfrac{\text{d}y}{\text{d}x} & = -3x^2y-y^3 \\ \dfrac{\text{d}y}{\text{d}x} & = \dfrac{-3x^2y-y^3}{x^3+3xy^2} \end{aligned}$$Gradien garis singgung diperoleh saat substitusi $x = 1$ dan $y = 2$ pada bentuk $\dfrac{\text{d}y}{\text{d}x}$ di atas, yaitu
$$\begin{aligned} m & = \left.\dfrac{\text{d}y}{\text{d}x}\right|_{(x,y) = (2,1)} \\ & = \dfrac{-3(1)^2(2)-2^3}{1^3+3(1)(2)^2} \\ & = \dfrac{-6-8}{1+12} = -\dfrac{14}{13}. \end{aligned}$$Persamaan garis singgung yang melalui titik $(x_1, y_1) = (1, 2)$ dan bergradien $m = -\dfrac{14}{13}$ dinyatakan oleh
$$\begin{aligned} y & = m(x-x_1)+y_1 \\ y & = -\dfrac{14}{13}(x-1) + 2 \\ 13y & = -14(x-1) + 26 \\ 13y+14x & = 40. \end{aligned}$$Jadi, persamaan garis singgung kurva $x^3y+xy^3 = 10$ di titik $(1, 2)$ adalah $$\boxed{13y + 14x = 40}.$$Berikut gambar grafik kurva dan garis singgungnya (menggunakan aplikasi GeoGebra).

Diketahui $x^2+y^2-6x+4y-12=0.$
Diferensialkan setiap suku terhadap variabel $x.$
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x} (x^2+y^2-6x+4y-12) & = \dfrac{\text{d}}{\text{d}x} (0) \\ 2x + 2y~\dfrac{\text{d}y}{\text{d}x}-6+4~\dfrac{\text{d}y}{\text{d}x}-0 & = 0 \\ (2y+4)~\dfrac{\text{d}y}{\text{d}x} & = 6-2x \\ \dfrac{\text{d}y}{\text{d}x} & = \dfrac{6-2x}{2y+4} \end{aligned}$$Karena titik singgungnya di $(7, 1),$ substitusikan $x = 7$ dan $y = 1$ pada $\dfrac{\text{d}y}{\text{d}x}$ di atas untuk memperoleh gradien garis.
$$\begin{aligned} m & = \left.\dfrac{\text{d}}{\text{d}x}\right|_{(x, y) = (7, 1)} \\ & = \dfrac{6-2(7)}{2(1)+4} \\ & = \dfrac{-8}{6} = -\dfrac43 \end{aligned}$$Persamaan garis yang melalui titik $(7, 1)$ dan bergradien $m = -\dfrac43$ adalah
$$\begin{aligned} y-y_1 & = m(x-x_1) \\ y-1 & = -\dfrac43(x-7) \\ 3y-3 & = -4x+28 \\ 4x+3y-31 & = 0. \end{aligned}$$Jadi, persamaan garis singgungnya adalah $\boxed{4x+3y-31=0}.$

Diketahui $\dfrac{x^2}{30} + \dfrac{y^2}{24} = 1.$
Diferensialkan setiap suku terhadap variabel $x.$
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x} \left(\dfrac{x^2}{30} + \dfrac{y^2}{24}\right) & = \dfrac{\text{d}}{\text{d}x} (1) \\ \dfrac{1}{30}(2x) + \dfrac{1}{24}(2y)~\dfrac{\text{d}y}{\text{d}x} & = 0 \\ \dfrac{y}{12}~\dfrac{\text{d}y}{\text{d}x} & = -\dfrac{x}{15} \\ \dfrac{\text{d}y}{\text{d}x} & = -\dfrac{4x}{5y} \end{aligned}$$Karena titik singgungnya di $(5, -2)$, substitusikan $x = 5$ dan $y = -2$ pada $\dfrac{\text{d}y}{\text{d}x}$ di atas untuk memperoleh gradien garis.
$$\begin{aligned} m & = \left.\dfrac{\text{d}}{\text{d}x}\right|_{(x, y) = (5, -2)} \\ & = -\dfrac{4\cancel{(5)}}{\cancel{5}(-2)} = 2 \end{aligned}$$Persamaan garis yang melalui titik $(5, -2)$ dan bergradien $m = 2$ adalah
$$\begin{aligned} y-y_1 & = m(x-x_1) \\ y-(-2) & = 2(x-5) \\ y+2 & = 2x-10 \\ y & = 2x-12. \end{aligned}$$Jadi, persamaan garis singgungnya adalah $\boxed{y=2x-12}.$

Diferensialkan setiap suku persamaan tersebut terhadap variabel $x.$
$$\begin{aligned} \dfrac{\text{d}}{\text{d}x} (y) + \dfrac{\text{d}}{\text{d}x} (\cos (xy^2)) + \dfrac{\text{d}}{\text{d}x} (3x^2) & = \dfrac{\text{d}y}{\text{d}x} (4) \\ \dfrac{\text{d}y}{\text{d}x} + \left[-y^2 \sin (xy^2)-2xy \sin (xy^2)~\dfrac{\text{d}y}{\text{d}x}\right] + 6x & = 0 \\ \left(1-2xy \sin (xy^2)\right)~\dfrac{\text{d}y}{\text{d}x} & = y^2 \sin (xy^2)-6x \\ \dfrac{\text{d}y}{\text{d}x} & = \dfrac{y^2 \sin (xy^2)-6x}{1-2xy \sin (xy^2)} \end{aligned}$$Gradien garis singgung diperoleh saat substitusi $x = 1$ dan $y = 0$ pada bentuk $\dfrac{\text{d}y}{\text{d}x}$ di atas, yaitu
$$\begin{aligned} m & = \left.\dfrac{\text{d}y}{\text{d}x}\right|_{(x,y) = (1,0)} \\ & = \dfrac{0-6(1)}{1-0} \\ & = \dfrac{-6}{1} = -6 \end{aligned}$$Persamaan garis singgung yang melalui titik $(x_1, y_1) = (1, 0)$ dan bergradien $m = -6$ dinyatakan oleh
$$\begin{aligned} y & = m(x-x_1)+y_1 \\ y & = -6(x-1) + 0 \\ y & = -6x + 6. \end{aligned}$$Jadi, persamaan garis singgung kurva $y + \cos (xy^2) + 3x^2 = 4$ di titik $(1,0)$ adalah $$\boxed{y = -6x + 6}.$$Berikut gambar grafik kurva dan garis singgungnya (menggunakan aplikasi GeoGebra).