derivative of ((x^4+1/(x^2+1))^5)

Lösung

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

\frac{5 ( x ^{4} + 1 ) ^{4} ( 2 x ^{5} + 4 x ^{3} - 2 x )}{( x ^{2} + 1 ) ^{6}}

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

5 (\frac{x ^{4} + 1}{x ^{2} + 1})^{4} \frac{2 x ^{5} + 4 x ^{3} - 2 x}{( x ^{2} + 1 ) ^{2}} : \frac{5 ( x ^{4} + 1 ) ^{4} ( 2 x ^{5} + 4 x ^{3} - 2 x )}{( x ^{2} + 1 ) ^{6}}

= \frac{5 ( x ^{4} + 1 ) ^{4} ( 2 x ^{5} + 4 x ^{3} - 2 x )}{( x ^{2} + 1 ) ^{6}}

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

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

\frac{\partial}{\partial x} (x^{2} + (y - 1)^{2})

\int_{- 4}^{4} \frac{1}{16 - x ^{2}} d x

\frac{\partial}{\partial x} (2 x + 1 + y e^{- x})