隱式微分計算器(附步驟)

逐步計算隱式導數

隱式微分計算器將求出隱式函數的一階與二階導數,可將 $$$y$$$ 視為 $$$x$$$ 的函數,或將 $$$x$$$ 視為 $$$y$$$ 的函數,並顯示步驟。

$$$($$$
,
$$$)$$$
若不需要在特定點處的導數,請留空。

如果計算器未能計算某些內容,或您發現了錯誤,或您有任何建議/回饋,請聯絡我們

您的輸入

$$$\frac{d}{dx} \left(x^{3} + y^{3} = 2 x y\right)$$$

解答

分別對等式兩邊求導(將 $$$y$$$ 視為 $$$x$$$ 的函數):$$$\frac{d}{dx} \left(x^{3} + y^{3}{\left(x \right)}\right) = \frac{d}{dx} \left(2 x y{\left(x \right)}\right)$$$

對等式左邊求導數。

和/差的導數等於導數的和/差:

$${\color{red}\left(\frac{d}{dx} \left(x^{3} + y^{3}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{3}\right) + \frac{d}{dx} \left(y^{3}{\left(x \right)}\right)\right)}$$

套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$,取 $$$n = 3$$$

$${\color{red}\left(\frac{d}{dx} \left(x^{3}\right)\right)} + \frac{d}{dx} \left(y^{3}{\left(x \right)}\right) = {\color{red}\left(3 x^{2}\right)} + \frac{d}{dx} \left(y^{3}{\left(x \right)}\right)$$

函數 $$$y^{3}{\left(x \right)}$$$ 是兩個函數 $$$f{\left(u \right)} = u^{3}$$$$$$g{\left(x \right)} = y{\left(x \right)}$$$ 之複合 $$$f{\left(g{\left(x \right)} \right)}$$$

應用鏈式法則 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$

$$3 x^{2} + {\color{red}\left(\frac{d}{dx} \left(y^{3}{\left(x \right)}\right)\right)} = 3 x^{2} + {\color{red}\left(\frac{d}{du} \left(u^{3}\right) \frac{d}{dx} \left(y{\left(x \right)}\right)\right)}$$

套用冪次法則 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$,取 $$$n = 3$$$

$$3 x^{2} + {\color{red}\left(\frac{d}{du} \left(u^{3}\right)\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) = 3 x^{2} + {\color{red}\left(3 u^{2}\right)} \frac{d}{dx} \left(y{\left(x \right)}\right)$$

返回原變數:

$$3 x^{2} + 3 {\color{red}\left(u\right)}^{2} \frac{d}{dx} \left(y{\left(x \right)}\right) = 3 x^{2} + 3 {\color{red}\left(y{\left(x \right)}\right)}^{2} \frac{d}{dx} \left(y{\left(x \right)}\right)$$

因此,$$$\frac{d}{dx} \left(x^{3} + y^{3}{\left(x \right)}\right) = 3 x^{2} + 3 y^{2}{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right)$$$

對等式右邊求導。

套用常數倍法則 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$,使用 $$$c = 2$$$$$$f{\left(x \right)} = x y{\left(x \right)}$$$

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

將乘積法則 $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ 應用於 $$$f{\left(x \right)} = x$$$$$$g{\left(x \right)} = y{\left(x \right)}$$$

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

套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$,取 $$$n = 1$$$,也就是 $$$\frac{d}{dx} \left(x\right) = 1$$$

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

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

因此,我們得到以下關於導數的線性方程:$$$3 x^{2} + 3 y^{2} \frac{dy}{dx} = 2 x \frac{dy}{dx} + 2 y$$$

解得 $$$\frac{dy}{dx} = \frac{3 x^{2} - 2 y}{2 x - 3 y^{2}}$$$

答案

$$$\frac{dy}{dx} = \frac{3 x^{2} - 2 y}{2 x - 3 y^{2}}$$$A


Please try a new game Rotatly