derivative y=x^2(4x^4+5x^3-16)

Lösung

ableitung von

y = x^{2} (4 x^{4} + 5 x^{3} - 16)

24 x^{5} + 25 x^{4} - 32 x

Schritte zur Lösung

\frac{d}{d x} (x^{2} (4 x^{4} + 5 x^{3} - 16))

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2}, g = 4 x^{4} + 5 x^{3} - 16

= \frac{d}{d x} (x^{2}) (4 x^{4} + 5 x^{3} - 16) + \frac{d}{d x} (4 x^{4} + 5 x^{3} - 16) x^{2}

\frac{d}{d x} (x^{2}) = 2 x

\frac{d}{d x} (4 x^{4} + 5 x^{3} - 16) = 16 x^{3} + 15 x^{2}

= 2 x (4 x^{4} + 5 x^{3} - 16) + (16 x^{3} + 15 x^{2}) x^{2}

Vereinfache

2 x (4 x^{4} + 5 x^{3} - 16) + (16 x^{3} + 15 x^{2}) x^{2} : 24 x^{5} + 25 x^{4} - 32 x

= 24 x^{5} + 25 x^{4} - 32 x

\frac{\partial}{\partial y} (1 + x^{3} + y^{3})

\int_{1}^{4} \frac{2}{x ^{3}} d x

d er i v a t i v e f (x) = 5 cot (\frac{x}{8})

x \to 0 lim (x^{5} ln (x))

\frac{d}{d y} (x^{y^{2}})