derivative f(x)=sqrt(3x^2+3x+4)

Lösung

ableitung von

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

\frac{6 x + 3}{2 3 x ^{2} + 3 x + 4}

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

\frac{d}{d x} (3 x^{2} + 3 x + 4) = 6 x + 3

= \frac{1}{2 3 x ^{2} + 3 x + 4} (6 x + 3)

Vereinfache

\frac{1}{2 3 x ^{2} + 3 x + 4} (6 x + 3) : \frac{6 x + 3}{2 3 x ^{2} + 3 x + 4}

= \frac{6 x + 3}{2 3 x ^{2} + 3 x + 4}

t an g e n t - x^{2} + 6 x

\frac{d}{d x} (\frac{8}{1 + x})

f (x) = ln (tan (x))

\frac{d x}{d t} = \frac{6 x ^{2} + t t ^{2} + x ^{2}}{6 t x}

\int_{0}^{3} (- x^{3} + 3 x^{2} - 2) d x