derivative of 3x^{k-2}+2kx^{k-1}+4x^k

Lösung

\frac{d}{d x} (3 x^{k - 2} + 2 k x^{k - 1} + 4 x^{k})

3 x^{k - 3} (k - 2) + 2 k x^{k - 2} (k - 1) + 4 k x^{k - 1}

Schritte zur Lösung

\frac{d}{d x} (3 x^{k - 2} + 2 k x^{k - 1} + 4 x^{k})

Wende die Summen-/Differenzregel an:

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

= \frac{d}{d x} (3 x^{k - 2}) + \frac{d}{d x} (2 k x^{k - 1}) + \frac{d}{d x} (4 x^{k})

\frac{d}{d x} (3 x^{k - 2}) = 3 x^{k - 3} (k - 2)

\frac{d}{d x} (2 k x^{k - 1}) = 2 k x^{k - 2} (k - 1)

\frac{d}{d x} (4 x^{k}) = 4 k x^{k - 1}

= 3 x^{k - 3} (k - 2) + 2 k x^{k - 2} (k - 1) + 4 k x^{k - 1}

\int \frac{x}{( x - 2 ) ^{3}} d x

\frac{d}{d x} (\frac{1}{4} e^{x} + \frac{3}{4} e^{- x})

\frac{d}{d x} (\frac{1 + x ^{2}}{x})

x \to 4 lim (\frac{6}{x})

\int x y^{2} d x