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

Lösung

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

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

Schritte zur Lösung

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

Wende Exponentenregel an:

\frac{1}{a} = a^{- 1}

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

\frac{\partial}{\partial x} (e^{- x a})

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = a, y^{^{'}} (0) = - 5

\int (t^{2} - 6)^{2} d t

d er i v a t i v e f (x) = \frac{x ^{2} + 5 x}{x + 2}

x \to \infty lim (x + 3)