(\partial)/(\partial x)((a^2x^2+b^2y^2)^4)

Lösung

\frac{\partial}{\partial x} ((a^{2} x^{2} + b^{2} y^{2})^{4})

8 a^{2} x (a^{2} x^{2} + b^{2} y^{2})^{3}

Schritte zur Lösung

\frac{\partial}{\partial x} ((a^{2} x^{2} + b^{2} y^{2})^{4})

Behandle

a, b, y

als Konstanten

Wende die Kettenregel an:

4 (a^{2} x^{2} + b^{2} y^{2})^{3} \frac{\partial}{\partial x} (a^{2} x^{2} + b^{2} y^{2})

= 4 (a^{2} x^{2} + b^{2} y^{2})^{3} \frac{\partial}{\partial x} (a^{2} x^{2} + b^{2} y^{2})

\frac{\partial}{\partial x} (a^{2} x^{2} + b^{2} y^{2}) = 2 a^{2} x

= 4 (a^{2} x^{2} + b^{2} y^{2})^{3} \cdot 2 a^{2} x

Vereinfache

= 8 a^{2} x (a^{2} x^{2} + b^{2} y^{2})^{3}

d er i v a t i v e f (x) = x^{2} - 2 x + 1

\frac{d}{d x} (\frac{x ^{3}}{3} + c)

x \to 5 lim (\frac{x ^{3} + 8}{x + 2})

x (x + 1) \frac{d y}{d x} + x y = 1, y (e) = 1

\frac{( x + x ) d y}{d x} = y + y