Turunan implisit dari $$$x^{2} y^{2} = 2 x + e^{y}$$$ terhadap $$$x$$$
Masukan Anda
Temukan $$$\frac{d}{dx} \left(x^{2} y^{2} = 2 x + e^{y}\right)$$$.
Solusi
Turunkan secara terpisah kedua ruas persamaan (anggap $$$y$$$ sebagai fungsi dari $$$x$$$): $$$\frac{d}{dx} \left(x^{2} y^{2}{\left(x \right)}\right) = \frac{d}{dx} \left(2 x + e^{y{\left(x \right)}}\right)$$$.
Turunkan ruas kiri dari persamaan.
Terapkan aturan hasil kali $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ pada $$$f{\left(x \right)} = x^{2}$$$ dan $$$g{\left(x \right)} = y^{2}{\left(x \right)}$$$:
$${\color{red}\left(\frac{d}{dx} \left(x^{2} y^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) y^{2}{\left(x \right)} + x^{2} \frac{d}{dx} \left(y^{2}{\left(x \right)}\right)\right)}$$Fungsi $$$y^{2}{\left(x \right)}$$$ merupakan komposisi $$$f{\left(g{\left(x \right)} \right)}$$$ dari dua fungsi $$$f{\left(u \right)} = u^{2}$$$ dan $$$g{\left(x \right)} = y{\left(x \right)}$$$.
Terapkan aturan rantai $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$$x^{2} {\color{red}\left(\frac{d}{dx} \left(y^{2}{\left(x \right)}\right)\right)} + y^{2}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right) = x^{2} {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(y{\left(x \right)}\right)\right)} + y^{2}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right)$$Terapkan aturan pangkat $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ dengan $$$n = 2$$$:
$$x^{2} {\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{2}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right) = x^{2} {\color{red}\left(2 u\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{2}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right)$$Kembalikan ke variabel semula:
$$2 x^{2} {\color{red}\left(u\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{2}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right) = 2 x^{2} {\color{red}\left(y{\left(x \right)}\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{2}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right)$$Terapkan aturan pangkat $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ dengan $$$n = 2$$$:
$$2 x^{2} y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{2}{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} = 2 x^{2} y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{2}{\left(x \right)} {\color{red}\left(2 x\right)}$$Sederhanakan:
$$2 x^{2} y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 x y^{2}{\left(x \right)} = 2 x \left(x \frac{d}{dx} \left(y{\left(x \right)}\right) + y{\left(x \right)}\right) y{\left(x \right)}$$Dengan demikian, $$$\frac{d}{dx} \left(x^{2} y^{2}{\left(x \right)}\right) = 2 x \left(x \frac{d}{dx} \left(y{\left(x \right)}\right) + y{\left(x \right)}\right) y{\left(x \right)}$$$.
Turunkan ruas kanan persamaan.
Turunan dari jumlah/selisih adalah jumlah/selisih dari turunan:
$${\color{red}\left(\frac{d}{dx} \left(2 x + e^{y{\left(x \right)}}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(2 x\right) + \frac{d}{dx} \left(e^{y{\left(x \right)}}\right)\right)}$$Fungsi $$$e^{y{\left(x \right)}}$$$ merupakan komposisi $$$f{\left(g{\left(x \right)} \right)}$$$ dari dua fungsi $$$f{\left(u \right)} = e^{u}$$$ dan $$$g{\left(x \right)} = y{\left(x \right)}$$$.
Terapkan aturan rantai $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(e^{y{\left(x \right)}}\right)\right)} + \frac{d}{dx} \left(2 x\right) = {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(y{\left(x \right)}\right)\right)} + \frac{d}{dx} \left(2 x\right)$$Turunan dari fungsi eksponensial adalah $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:
$${\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + \frac{d}{dx} \left(2 x\right) = {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + \frac{d}{dx} \left(2 x\right)$$Kembalikan ke variabel semula:
$$e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + \frac{d}{dx} \left(2 x\right) = e^{{\color{red}\left(y{\left(x \right)}\right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + \frac{d}{dx} \left(2 x\right)$$Terapkan aturan kelipatan konstanta $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ dengan $$$c = 2$$$ dan $$$f{\left(x \right)} = x$$$:
$$e^{y{\left(x \right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + {\color{red}\left(\frac{d}{dx} \left(2 x\right)\right)} = e^{y{\left(x \right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + {\color{red}\left(2 \frac{d}{dx} \left(x\right)\right)}$$Terapkan aturan pangkat $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ dengan $$$n = 1$$$, dengan kata lain, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$e^{y{\left(x \right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = e^{y{\left(x \right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 {\color{red}\left(1\right)}$$Dengan demikian, $$$\frac{d}{dx} \left(2 x + e^{y{\left(x \right)}}\right) = e^{y{\left(x \right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2$$$.
Dengan demikian, kita memperoleh persamaan linier berikut terhadap turunan: $$$2 x y \left(x \frac{dy}{dx} + y\right) = e^{y} \frac{dy}{dx} + 2$$$.
Dengan menyelesaikannya, kita memperoleh bahwa $$$\frac{dy}{dx} = \frac{- 2 x y^{2} + 2}{2 x^{2} y - e^{y}}$$$.
Jawaban
$$$\frac{dy}{dx} = \frac{- 2 x y^{2} + 2}{2 x^{2} y - e^{y}}$$$A