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

Lösung

ableitung von

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

16 x^{3} - 15 x^{2} + 2

Schritte zur Lösung

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

Wende die Summen-/Differenzregel an:

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

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

\frac{d}{d x} (4 x^{4}) = 16 x^{3}

\frac{d}{d x} (5 x^{3}) = 15 x^{2}

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

\frac{d}{d x} (3) = 0

= 16 x^{3} - 15 x^{2} + 2 - 0

Vereinfache

= 16 x^{3} - 15 x^{2} + 2

\frac{d}{d x} (\frac{x}{a ^{2} a ^{2} + x ^{2}})

\frac{d}{d x} (25 ln (x^{2}) - x^{2})

\int_{1}^{4} x d x

\int e^{- 5 x} x d x

im pl i c i t \frac{d y}{d x}, x^{\frac{2}{3}} + y^{\frac{2}{3}} = 3