derivative of 2x^{4+7x+5}

Lösung

\frac{d}{d x} (2 x^{4 + 7 x + 5})

2 x^{7 x + 9} (7 ln (x) + \frac{7 x + 9}{x})

Schritte zur Lösung

\frac{d}{d x} (2 x^{4 + 7 x + 5})

Vereinfache

2 x^{4 + 7 x + 5} : 2 x^{7 x + 9}

= \frac{d}{d x} (2 x^{7 x + 9})

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d x} (x^{7 x + 9})

Wende Exponentenregel an:

a^{b} = e^{b l n (a)}

x^{7 x + 9} = e^{(7 x + 9) l n (x)}

= 2 \frac{d}{d x} (e^{(7 x + 9) l n (x)})

Wende die Kettenregel an:

e^{(7 x + 9) l n (x)} \frac{d}{d x} ((7 x + 9) ln (x))

= e^{(7 x + 9) l n (x)} \frac{d}{d x} ((7 x + 9) ln (x))

\frac{d}{d x} ((7 x + 9) ln (x)) = 7 ln (x) + \frac{7 x + 9}{x}

= 2 e^{(7 x + 9) l n (x)} (7 ln (x) + \frac{7 x + 9}{x})

e^{(7 x + 9) l n (x)} = x^{7 x + 9}

= 2 x^{7 x + 9} (7 ln (x) + \frac{7 x + 9}{x})

\int \frac{y}{( 1 + y ^{2} )} d y

\int \frac{4 x ^{2} - 3 x + 4}{x ^{3} + 4 x} d x

y^{^{''}} + 2 y^{^{'}} + y = 2 e^{- x}

\frac{d}{d x} (36 x^{2} - 72 x)

x \to 0 lim (cos (\frac{π}{x}))