derivative f(x)=(\sqrt[3]{4x^4+5x^3})

Lösung

ableitung von

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

\frac{16 x ^{3} + 15 x ^{2}}{3 ( x ^{3} ) ^{\frac{2}{3}} ( 4 x + 5 ) ^{\frac{2}{3}}}

Schritte zur Lösung

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

Wende die Kettenregel an:

\frac{1}{3 ( 4 x ^{4} + 5 x ^{3} ) ^{\frac{2}{3}}} \frac{d}{d x} (4 x^{4} + 5 x^{3})

= \frac{1}{3 ( 4 x ^{4} + 5 x ^{3} ) ^{\frac{2}{3}}} \frac{d}{d x} (4 x^{4} + 5 x^{3})

\frac{d}{d x} (4 x^{4} + 5 x^{3}) = 16 x^{3} + 15 x^{2}

= \frac{1}{3 ( 4 x ^{4} + 5 x ^{3} ) ^{\frac{2}{3}}} (16 x^{3} + 15 x^{2})

Vereinfache

\frac{1}{3 ( 4 x ^{4} + 5 x ^{3} ) ^{\frac{2}{3}}} (16 x^{3} + 15 x^{2}) : \frac{16 x ^{3} + 15 x ^{2}}{3 ( x ^{3} ) ^{\frac{2}{3}} ( 4 x + 5 ) ^{\frac{2}{3}}}

= \frac{16 x ^{3} + 15 x ^{2}}{3 ( x ^{3} ) ^{\frac{2}{3}} ( 4 x + 5 ) ^{\frac{2}{3}}}

\int_{2}^{4} \frac{6 - x - y}{8} d y

in v erse l a pl a ce \frac{2}{s + 1} + \frac{8 s}{s ^{2} + 9}

x \to 0 lim (\frac{e}{x})

\int \frac{1}{( x + y )} d x

d er i v a t i v e (x + 1)^{3}