d/(da)(a^5+a^2b^3-a^3b^2-b^5)

Lösung

\frac{d}{d a} (a^{5} + a^{2} b^{3} - a^{3} b^{2} - b^{5})

5 a^{4} - 3 a^{2} b^{2} + 2 a b^{3}

Schritte zur Lösung

\frac{d}{d a} (a^{5} + a^{2} b^{3} - a^{3} b^{2} - b^{5})

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d a} (a^{5}) + \frac{d}{d a} (a^{2} b^{3}) - \frac{d}{d a} (a^{3} b^{2}) - \frac{d}{d a} (b^{5})

\frac{d}{d a} (a^{5}) = 5 a^{4}

\frac{d}{d a} (a^{2} b^{3}) = 2 b^{3} a

\frac{d}{d a} (a^{3} b^{2}) = 3 b^{2} a^{2}

\frac{d}{d a} (b^{5}) = 0

= 5 a^{4} + 2 b^{3} a - 3 b^{2} a^{2} - 0

Vereinfache

= 5 a^{4} - 3 a^{2} b^{2} + 2 a b^{3}

e x t re m e f (x) = \frac{2}{3} x^{3} - \frac{7}{2} x^{2} + 3 x + 16

e x t re m e x ln (3 x)

e x t re m e f (x) = \frac{1}{x} - ln (x)

e x t re m e f (x) = - 2 x^{3} + 42 x^{2} - 240 x + 5

e x t re m e y = 6 x - x^{2} - 2