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

Lösung

ableitung von

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

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

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

\int e^{x} (1 + e^{1 - x}) d x

x y^{^{'}} + 6 x^{2} y = x e^{5 x}

\int sec (x) tan^{2} (x) d x

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

in v erse l a pl a ce \frac{12}{s}