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Implizite Ableitung von $$$x^{2} y^{2} = 2 x + e^{y}$$$ nach $$$x$$$
Ihre Eingabe
Bestimme $$$\frac{d}{dx} \left(x^{2} y^{2} = 2 x + e^{y}\right)$$$.
Lösung
Leite beide Seiten der Gleichung getrennt ab (betrachte $$$y$$$ als Funktion von $$$x$$$): $$$\frac{d}{dx} \left(x^{2} y^{2}{\left(x \right)}\right) = \frac{d}{dx} \left(2 x + e^{y{\left(x \right)}}\right)$$$.
Leite die linke Seite der Gleichung ab.
Wende die Produktregel $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ mit $$$f{\left(x \right)} = x^{2}$$$ und $$$g{\left(x \right)} = y^{2}{\left(x \right)}$$$ an:
$${\color{red}\left(\frac{d}{dx} \left(x^{2} y^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) y^{2}{\left(x \right)} + x^{2} \frac{d}{dx} \left(y^{2}{\left(x \right)}\right)\right)}$$Wende die Potenzregel $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ mit $$$n = 2$$$ an:
$$x^{2} \frac{d}{dx} \left(y^{2}{\left(x \right)}\right) + y^{2}{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} = x^{2} \frac{d}{dx} \left(y^{2}{\left(x \right)}\right) + y^{2}{\left(x \right)} {\color{red}\left(2 x\right)}$$Die Funktion $$$y^{2}{\left(x \right)}$$$ ist die Komposition $$$f{\left(g{\left(x \right)} \right)}$$$ der beiden Funktionen $$$f{\left(u \right)} = u^{2}$$$ und $$$g{\left(x \right)} = y{\left(x \right)}$$$.
Wende die Kettenregel $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ an:
$$x^{2} {\color{red}\left(\frac{d}{dx} \left(y^{2}{\left(x \right)}\right)\right)} + 2 x y^{2}{\left(x \right)} = x^{2} {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(y{\left(x \right)}\right)\right)} + 2 x y^{2}{\left(x \right)}$$Wende die Potenzregel $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ mit $$$n = 2$$$ an:
$$x^{2} {\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 x y^{2}{\left(x \right)} = x^{2} {\color{red}\left(2 u\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 x y^{2}{\left(x \right)}$$Zurück zur ursprünglichen Variable:
$$2 x^{2} {\color{red}\left(u\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 x y^{2}{\left(x \right)} = 2 x^{2} {\color{red}\left(y{\left(x \right)}\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 x y^{2}{\left(x \right)}$$Vereinfachen:
$$2 x^{2} y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 x y^{2}{\left(x \right)} = 2 x \left(x \frac{d}{dx} \left(y{\left(x \right)}\right) + y{\left(x \right)}\right) y{\left(x \right)}$$Somit gilt $$$\frac{d}{dx} \left(x^{2} y^{2}{\left(x \right)}\right) = 2 x \left(x \frac{d}{dx} \left(y{\left(x \right)}\right) + y{\left(x \right)}\right) y{\left(x \right)}$$$.
Leite die rechte Seite der Gleichung ab.
Die Ableitung einer Summe/Differenz ist die Summe/Differenz der Ableitungen:
$${\color{red}\left(\frac{d}{dx} \left(2 x + e^{y{\left(x \right)}}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(2 x\right) + \frac{d}{dx} \left(e^{y{\left(x \right)}}\right)\right)}$$Wende die Konstantenfaktorregel $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ mit $$$c = 2$$$ und $$$f{\left(x \right)} = x$$$ an:
$${\color{red}\left(\frac{d}{dx} \left(2 x\right)\right)} + \frac{d}{dx} \left(e^{y{\left(x \right)}}\right) = {\color{red}\left(2 \frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(e^{y{\left(x \right)}}\right)$$Wenden Sie die Potenzregel $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ mit $$$n = 1$$$ an, mit anderen Worten, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$2 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(e^{y{\left(x \right)}}\right) = 2 {\color{red}\left(1\right)} + \frac{d}{dx} \left(e^{y{\left(x \right)}}\right)$$Die Funktion $$$e^{y{\left(x \right)}}$$$ ist die Komposition $$$f{\left(g{\left(x \right)} \right)}$$$ der beiden Funktionen $$$f{\left(u \right)} = e^{u}$$$ und $$$g{\left(x \right)} = y{\left(x \right)}$$$.
Wende die Kettenregel $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ an:
$${\color{red}\left(\frac{d}{dx} \left(e^{y{\left(x \right)}}\right)\right)} + 2 = {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(y{\left(x \right)}\right)\right)} + 2$$Die Ableitung der Exponentialfunktion ist $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:
$${\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 = {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2$$Zurück zur ursprünglichen Variable:
$$e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 = e^{{\color{red}\left(y{\left(x \right)}\right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2$$Somit gilt $$$\frac{d}{dx} \left(2 x + e^{y{\left(x \right)}}\right) = e^{y{\left(x \right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2$$$.
Daher haben wir die folgende lineare Gleichung bezüglich der Ableitung erhalten: $$$2 x y \left(x \frac{dy}{dx} + y\right) = e^{y} \frac{dy}{dx} + 2$$$.
Beim Lösen ergibt sich, dass $$$\frac{dy}{dx} = \frac{- 2 x y^{2} + 2}{2 x^{2} y - e^{y}}$$$.
Antwort
$$$\frac{dy}{dx} = \frac{- 2 x y^{2} + 2}{2 x^{2} y - e^{y}}$$$A