Tag: Aturan Rantai

Diferensialkan setiap suku terhadap $x.$
$$\begin{array}{rl}{\displaystyle \frac{\text{d}}{\text{d}x}}(a{x}^2)+{\displaystyle \frac{\text{d}}{\text{d}x}}(b{y}^2)& ={\displaystyle \frac{\text{d}}{\text{d}x}}(1)\\ 2ax+(0+2by{\displaystyle \frac{\text{d}y}{\text{d}x}})& =0\\ 2by{\displaystyle \frac{\text{d}y}{\text{d}x}}& =-2ax\\ {\displaystyle \frac{\text{d}y}{\text{d}x}}& ={\displaystyle \frac{-2ax}{2by}}\\ {\displaystyle \frac{\text{d}y}{\text{d}x}}& =-{\displaystyle \frac{ax}{by}}\end{array}$$Jadi, diperoleh $\boxed{{\displaystyle {\displaystyle \frac{\text{d}y}{\text{d}x}}=-{\displaystyle \frac{ax}{by}}}}.$

Diferensialkan setiap suku terhadap $x.$
$$\begin{array}{rl}{\displaystyle \frac{\text{d}}{\text{d}x}}(\sqrt{x})+{\displaystyle \frac{\text{d}}{\text{d}x}}(\sqrt{y})& ={\displaystyle \frac{\text{d}}{\text{d}x}}(a)\\ {\displaystyle \frac{1}{2\sqrt{x}}}+(0+{\displaystyle \frac{1}{2\sqrt{y}}}{\displaystyle \frac{\text{d}y}{\text{d}x}})& =0\\ {\displaystyle \frac{1}{2\sqrt{y}}}{\displaystyle \frac{\text{d}y}{\text{d}x}}& =-{\displaystyle \frac{1}{2\sqrt{x}}}\\ {\displaystyle \frac{\text{d}y}{\text{d}x}}& =-{\displaystyle \frac{1}{2\sqrt{x}}}\cdot 2\sqrt{y}\\ {\displaystyle \frac{\text{d}y}{\text{d}x}}& =-\sqrt{{\displaystyle \frac{y}{x}}}\end{array}$$Jadi, diperoleh $\boxed{{\displaystyle {\displaystyle \frac{\text{d}y}{\text{d}x}}=-\sqrt{{\displaystyle \frac{y}{x}}}}}.$

Diferensialkan setiap suku terhadap $x.$
$$\begin{array}{rl}{\displaystyle \frac{\text{d}}{\text{d}x}}(\sin x)+{\displaystyle \frac{\text{d}}{\text{d}x}}(\sin y)& ={\displaystyle \frac{\text{d}}{\text{d}x}}(\pi )\\ \cos x+(0+\cos y{\displaystyle \frac{\text{d}y}{\text{d}x}})& =0\\ \cos y{\displaystyle \frac{\text{d}y}{\text{d}x}}& =-\cos x\\ {\displaystyle \frac{\text{d}y}{\text{d}x}}& =-{\displaystyle \frac{\cos x}{\cos y}}\end{array}$$Jadi, diperoleh $\boxed{{\displaystyle {\displaystyle \frac{\text{d}y}{\text{d}x}}=-{\displaystyle \frac{\cos x}{\cos y}}}}.$

Perhatikan bahwa $3xy-4=x$ ekuivalen dengan $y={\displaystyle \frac{x+4}{3x}}$. Sekarang, $y$ dinyatakan sebagai fungsi eksplisit dari $x$ dan akan kita cari turunannya menggunakan aturan hasil bagi.
Misalkan
$$\begin{array}{rl}u& =x+4\Rightarrow u^{\prime }=1.\\ v& =3x\Rightarrow v^{\prime }=3.\end{array}$$Kita peroleh
$$\begin{array}{rl}{\displaystyle \frac{\text{d}y}{\text{d}x}}& ={\displaystyle \frac{u^{\prime }v-u{v}^{\prime }}{v^2}}\\ & ={\displaystyle \frac{1(3x)-(x+4)(3)}{(3x{)}^2}}\\ & ={\displaystyle \frac{3x-3x-12}{9x^2}}\\ & ={\displaystyle \frac{-12}{9x^2}}=-{\displaystyle \frac{4}{3x^2}}.\end{array}$$Sekarang kita akan menurunkan secara implisit fungsi $3xy-4=x.$
Diferensialkan setiap suku terhadap $x.$
$$\begin{array}{rl}{\displaystyle \frac{\text{d}}{\text{d}x}}(3xy)-{\displaystyle \frac{\text{d}}{\text{d}x}}(4)& ={\displaystyle \frac{\text{d}}{\text{d}x}}(x)\\ (3y+3x{\displaystyle \frac{\text{d}y}{\text{d}x}})-0& =1\\ 3x{\displaystyle \frac{\text{d}y}{\text{d}x}}& =1-3y\\ {\displaystyle \frac{\text{d}y}{\text{d}x}}& ={\displaystyle \frac{1-3y}{3x}}\end{array}$$Sekarang kita peroleh turunan implisitnya, tetapi perhatikan bahwa ruas kanannya masih memuat variabel $y.$ Oleh karena itu, kita perlu lakukan substitusi $y={\displaystyle \frac{x+4}{3x}}.$
$$\begin{array}{rl}{\displaystyle \frac{\text{d}y}{\text{d}x}}& ={\displaystyle \frac{1-3\left({\displaystyle \frac{x+4}{3x}}\right)}{3x}}\\ & ={\displaystyle \frac{{\displaystyle \frac{3x}{3x}}-{\displaystyle \frac{3x+12}{3x}}}{3x}}\\ & ={\displaystyle \frac{-12}{9x^2}}=-{\displaystyle \frac{4}{3x^2}}\end{array}$$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
$$\begin{array}{rl}(x^2-6x+9)y& =4\\ (x-3{)}^{2}y& =4\\ y& ={\displaystyle \frac{4}{(x-3{)}^2}}.\end{array}$$Berikutnya, $y$ dinyatakan sebagai fungsi eksplisit dari $x$ dan akan kita cari turunannya menggunakan aturan basil bagi.
Misalkan
$$\begin{array}{rl}u& =4\Rightarrow u^{\prime }=0.\\ v& =(x-3{)}^2\Rightarrow v^{\prime }=2(x-3)=2x-6.\end{array}$$Kita peroleh
$$\begin{array}{rl}{\displaystyle \frac{\text{d}y}{\text{d}x}}& ={\displaystyle \frac{u^{\prime }v-u{v}^{\prime }}{v^2}}\\ & ={\displaystyle \frac{0(x-3{)}^2-4(2x-6)}{{((x-3{)}^2))}^2}}\\ & ={\displaystyle \frac{-8x+24}{(x-3{)}^4}}.\end{array}$$Kita akan menurunkan secara implisit fungsi $x^{2}y-6xy+9y=4.$
Diferensialkan setiap suku terhadap $x.$
$$\begin{array}{rl}{\displaystyle \frac{\text{d}}{\text{d}x}}(x^{2}y)-{\displaystyle \frac{\text{d}}{\text{d}x}}(6xy)+{\displaystyle \frac{\text{d}}{\text{d}x}}(9y)& ={\displaystyle \frac{\text{d}}{\text{d}x}}(4)\\ (2xy+x^2{\displaystyle \frac{\text{d}y}{\text{d}x}})-6(y+x{\displaystyle \frac{\text{d}y}{\text{d}x}})+9{\displaystyle \frac{\text{d}y}{\text{d}x}}& =0\\ (x^2-6x+9){\displaystyle \frac{\text{d}y}{\text{d}x}}& =-2xy+6y\\ (x-3{)}^2{\displaystyle \frac{\text{d}y}{\text{d}x}}& =(-2x+6)y\\ {\displaystyle \frac{\text{d}y}{\text{d}x}}& ={\displaystyle \frac{(-2x+6)y}{(x-3{)}^2}}\end{array}$$Kita peroleh turunan implisitnya, tetapi perhatikan bahwa ruas kanannya masih memuat variabel $y$. Oleh karena itu, kita perlu lakukan substitusi $y={\displaystyle \frac{4}{(x-3{)}^2}}.$
$$\begin{array}{rl}{\displaystyle \frac{\text{d}y}{\text{d}x}}& ={\displaystyle \frac{(-2x+6)\left({\displaystyle \frac{4}{(x-3{)}^2}}\right)}{(x-3{)}^2}}\\ & ={\displaystyle \frac{-8x+24}{(x-3{)}^4}}\end{array}$$Jadi, terbukti bahwa fungsi $x^{2}y-6xy+9y=4$ 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
$$\begin{array}{rl}{f}^{\prime }(x,y)& =8(x+2y{)}^7(x{\displaystyle \frac{\text{d}}{\text{d}x}}+2y{\displaystyle \frac{\text{d}}{\text{d}y}}{\displaystyle \frac{\text{d}y}{\text{d}x}})\\ & =8(x+2y{)}^7(1+2{\displaystyle \frac{\text{d}y}{\text{d}x}})\\ & =8(x+2y{)}^7+16(x+2y{)}^7{\displaystyle \frac{\text{d}y}{\text{d}x}}\end{array}$$Jadi, turunan pertama fungsi implisit itu adalah $\boxed{{\displaystyle 8(x+2y{)}^7+16(x+2y{)}^7{\displaystyle \frac{\text{d}y}{\text{d}x}}}}.$

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

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

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

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

Fungsi $f$ dinyatakan dalam bentuk pecahan sehingga kita dapat menggunakan aturan hasil bagi untuk menentukan turunannya.
Misalkan
$$\begin{array}{rl}u& =y-x^2\Rightarrow u^{\prime }={\displaystyle \frac{\text{d}y}{\text{d}x}}-2x\\ v& =y^2-x\Rightarrow v^{\prime }=2y{\displaystyle \frac{\text{d}y}{\text{d}x}}-1.\end{array}$$Dengan demikian, kita peroleh
$$\begin{array}{rl}{f}^{\prime }(x,y)& ={\displaystyle \frac{u^{\prime }v-u{v}^{\prime }}{v^2}}\\ & ={\displaystyle \frac{({\displaystyle \frac{\text{d}y}{\text{d}x}}-2x)(y^2-x)-(y-x^2)(2y{\displaystyle \frac{\text{d}y}{\text{d}x}}-1)}{(y^2-x{)}^2}}\\ & ={\displaystyle \frac{(y^2-x-2y^2+2x^{2}y){\displaystyle \frac{\text{d}y}{\text{d}x}}-2x{y}^2+2x^2+y-x^2}{(y^2-x{)}^2}}\\ & ={\displaystyle \frac{(-x-y^2+2x^{2}y){\displaystyle \frac{\text{d}y}{\text{d}x}}-2x{y}^2+x^2+y}{(y^2-x{)}^2}}.\end{array}$$Jadi, turunan pertama fungsi implisit $f(x,y)={\displaystyle \frac{y-x^2}{y^2-x}}$ terhadap variabel $x$ adalah $$\boxed{{\displaystyle f^{\prime }(x,y)={\displaystyle \frac{(-x-y^2+2x^{2}y){\displaystyle \frac{\text{d}y}{\text{d}x}}-2x{y}^2+x^2+y}{(y^2-x{)}^2}}}}.$$

Diferensialkan setiap suku terhadap $x.$
$$\begin{array}{rl}{\displaystyle \frac{\text{d}}{\text{d}x}}(x^3)-{\displaystyle \frac{\text{d}}{\text{d}x}}(xy)+{\displaystyle \frac{\text{d}}{\text{d}x}}(y^3)& ={\displaystyle \frac{\text{d}}{\text{d}x}}(1)\\ 3x^2-(x{\displaystyle \frac{\text{d}y}{\text{d}x}}+y)+3y^2{\displaystyle \frac{\text{d}y}{\text{d}x}}& =0\\ (-x+3y^2){\displaystyle \frac{\text{d}y}{\text{d}x}}+3x^2-y& =0\\ {\displaystyle \frac{\text{d}y}{\text{d}x}}& ={\displaystyle \frac{y-3x^2}{3y^2-x}}\end{array}$$Jadi, diperoleh $\boxed{{\displaystyle {\displaystyle \frac{\text{d}y}{\text{d}x}}={\displaystyle \frac{y-3x^2}{3y^2-x}}}}.$

Diferensialkan setiap suku terhadap $x.$
$$\begin{array}{rl}{\displaystyle \frac{\text{d}}{\text{d}x}}(3x+7{)}^5& ={\displaystyle \frac{\text{d}}{\text{d}x}}(2y^3)\\ 5(3x+7{)}^4\cdot \underset{\text{Turunan}(3x+7)}{\underbrace{3}}& =6y^2{\displaystyle \frac{\text{d}y}{\text{d}x}}\\ {\displaystyle \frac{\text{d}y}{\text{d}x}}& ={\displaystyle \frac{5(3x+7{)}^4}{2y^2}}\end{array}$$Jadi, diperoleh $\boxed{{\displaystyle {\displaystyle \frac{\text{d}y}{\text{d}x}}={\displaystyle \frac{5(3x+7{)}^4}{2y^2}}}}.$

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

Pertama, kita dapat tuliskan persamaan di atas dalam bentuk
$$\begin{array}{rl}{y}^3(x+y)& =x-y\\ x{y}^3+y^4& =x-y.\end{array}$$Selanjutnya, diferensialkan terhadap variabel $x.$
$$\begin{array}{rl}{\displaystyle \frac{\text{d}}{\text{d}x}}(x{y}^3)+{\displaystyle \frac{\text{d}}{\text{d}x}}(y^4)& ={\displaystyle \frac{\text{d}}{\text{d}x}}(x)-{\displaystyle \frac{\text{d}}{\text{d}x}}(y)\\ (3x{y}^2{\displaystyle \frac{\text{d}y}{\text{d}x}}+y^3)+4y^3{\displaystyle \frac{\text{d}y}{\text{d}x}}& =1-{\displaystyle \frac{\text{d}y}{\text{d}x}}\\ (3x{y}^2+4y^3+1){\displaystyle \frac{\text{d}y}{\text{d}x}}& =1-y^3\\ {\displaystyle \frac{\text{d}y}{\text{d}x}}& ={\displaystyle \frac{1-y^3}{1+3x{y}^2+4y^3}}\end{array}$$Jadi, diperoleh $\boxed{{\displaystyle {\displaystyle \frac{\text{d}y}{\text{d}x}}={\displaystyle \frac{1-y^3}{1+3x{y}^2+4y^3}}}}.$ 

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

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

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

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