$$$x$$$에 대한 $$$x^{2} y^{2} = 2 x + e^{y}$$$의 암시적 도함수

방정식의 양변을 각각 미분하시오($$$y$$$를 $$$x$$$의 함수로 보고): $$$\frac{d}{dx} \left(x^{2} y^{2}{\left(x \right)}\right) = \frac{d}{dx} \left(2 x + e^{y{\left(x \right)}}\right)$$$.

방정식의 좌변을 미분하세요.

$$$f{\left(x \right)} = x^{2}$$$와 $$$g{\left(x \right)} = y^{2}{\left(x \right)}$$$에 대해 곱의 미분법칙 $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$을 적용하십시오:

$${\color{red}\left(\frac{d}{dx} \left(x^{2} y^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) y^{2}{\left(x \right)} + x^{2} \frac{d}{dx} \left(y^{2}{\left(x \right)}\right)\right)}$$

함수 $$$y^{2}{\left(x \right)}$$$는 두 함수 $$$f{\left(u \right)} = u^{2}$$$와 $$$g{\left(x \right)} = y{\left(x \right)}$$$의 합성함수 $$$f{\left(g{\left(x \right)} \right)}$$$이다.

연쇄법칙 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$을(를) 적용하십시오:

$$x^{2} {\color{red}\left(\frac{d}{dx} \left(y^{2}{\left(x \right)}\right)\right)} + y^{2}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right) = x^{2} {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(y{\left(x \right)}\right)\right)} + y^{2}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right)$$

거듭제곱법칙 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$을 $$$n = 2$$$에 적용합니다:

$$x^{2} {\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{2}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right) = x^{2} {\color{red}\left(2 u\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{2}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right)$$

역치환:

$$2 x^{2} {\color{red}\left(u\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{2}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right) = 2 x^{2} {\color{red}\left(y{\left(x \right)}\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{2}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right)$$

거듭제곱법칙 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$을 $$$n = 2$$$에 적용합니다:

$$2 x^{2} y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{2}{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} = 2 x^{2} y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{2}{\left(x \right)} {\color{red}\left(2 x\right)}$$

간단히 하시오:

$$2 x^{2} y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 x y^{2}{\left(x \right)} = 2 x \left(x \frac{d}{dx} \left(y{\left(x \right)}\right) + y{\left(x \right)}\right) y{\left(x \right)}$$

따라서, $$$\frac{d}{dx} \left(x^{2} y^{2}{\left(x \right)}\right) = 2 x \left(x \frac{d}{dx} \left(y{\left(x \right)}\right) + y{\left(x \right)}\right) y{\left(x \right)}$$$.

방정식의 우변을 미분하시오.

합/차의 도함수는 도함수들의 합/차이다:

$${\color{red}\left(\frac{d}{dx} \left(2 x + e^{y{\left(x \right)}}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(2 x\right) + \frac{d}{dx} \left(e^{y{\left(x \right)}}\right)\right)}$$

함수 $$$e^{y{\left(x \right)}}$$$는 두 함수 $$$f{\left(u \right)} = e^{u}$$$와 $$$g{\left(x \right)} = y{\left(x \right)}$$$의 합성함수 $$$f{\left(g{\left(x \right)} \right)}$$$이다.

연쇄법칙 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$을(를) 적용하십시오:

$${\color{red}\left(\frac{d}{dx} \left(e^{y{\left(x \right)}}\right)\right)} + \frac{d}{dx} \left(2 x\right) = {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(y{\left(x \right)}\right)\right)} + \frac{d}{dx} \left(2 x\right)$$

지수함수의 도함수는 $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:

$${\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + \frac{d}{dx} \left(2 x\right) = {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + \frac{d}{dx} \left(2 x\right)$$

역치환:

$$e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + \frac{d}{dx} \left(2 x\right) = e^{{\color{red}\left(y{\left(x \right)}\right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + \frac{d}{dx} \left(2 x\right)$$

상수배 법칙 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$을 $$$c = 2$$$와 $$$f{\left(x \right)} = x$$$에 적용합니다:

$$e^{y{\left(x \right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + {\color{red}\left(\frac{d}{dx} \left(2 x\right)\right)} = e^{y{\left(x \right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + {\color{red}\left(2 \frac{d}{dx} \left(x\right)\right)}$$

멱법칙 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$을 $$$n = 1$$$에 대해 적용하면, 즉 $$$\frac{d}{dx} \left(x\right) = 1$$$:

$$e^{y{\left(x \right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = e^{y{\left(x \right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 {\color{red}\left(1\right)}$$

따라서, $$$\frac{d}{dx} \left(2 x + e^{y{\left(x \right)}}\right) = e^{y{\left(x \right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2$$$.

따라서 도함수에 대한 다음과 같은 선형 방정식을 얻었다: $$$2 x y \left(x \frac{dy}{dx} + y\right) = e^{y} \frac{dy}{dx} + 2$$$

이를 풀면 $$$\frac{dy}{dx} = \frac{- 2 x y^{2} + 2}{2 x^{2} y - e^{y}}$$$라는 결과를 얻습니다.