derivative of (((x^2+2x)/(x+1))^5)

Lösung

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

\frac{5 ( x ^{2} + 2 x ) ^{4} ( x ^{2} + 2 x + 2 )}{( x + 1 ) ^{6}}

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

5 (\frac{x ^{2} + 2 x}{x + 1})^{4} \frac{x ^{2} + 2 x + 2}{( x + 1 ) ^{2}} : \frac{5 ( x ^{2} + 2 x ) ^{4} ( x ^{2} + 2 x + 2 )}{( x + 1 ) ^{6}}

= \frac{5 ( x ^{2} + 2 x ) ^{4} ( x ^{2} + 2 x + 2 )}{( x + 1 ) ^{6}}

\frac{\partial}{\partial x} (y arcsin (\frac{x}{y}))

x \to - \infty lim (\frac{4 x ^{4} - 3 + 6 x}{1 - 2 x})

\int p^{3} ln (p) d p

a re a y = x^{4}, y = x^{2}

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