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

Lösung

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

\frac{10 ( 2 x + 1 ) ^{4} ( x ^{2} + 1 ) - x ( 2 x + 1 ) ^{5}}{( x ^{2} + 1 ) x ^{2} + 1}

Schritte zur Lösung

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

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} ((2 x + 1)^{5}) = 10 (2 x + 1)^{4}

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

= \frac{10 ( 2 x + 1 ) ^{4} x ^{2} + 1 - \frac{x}{x ^{2} + 1} ( 2 x + 1 ) ^{5}}{( x ^{2} + 1 ) ^{2}}

Vereinfache

\frac{10 ( 2 x + 1 ) ^{4} x ^{2} + 1 - \frac{x}{x ^{2} + 1} ( 2 x + 1 ) ^{5}}{( x ^{2} + 1 ) ^{2}} : \frac{10 ( 2 x + 1 ) ^{4} ( x ^{2} + 1 ) - x ( 2 x + 1 ) ^{5}}{( x ^{2} + 1 ) x ^{2} + 1}

= \frac{10 ( 2 x + 1 ) ^{4} ( x ^{2} + 1 ) - x ( 2 x + 1 ) ^{5}}{( x ^{2} + 1 ) x ^{2} + 1}

t an g e n t f (x) = x + \frac{4}{x}, at x = 4

\frac{d}{d z} (ln (\frac{c ^{2} - z ^{2}}{c ^{2} + z ^{2}}))

y^{^{''}} + 4 y^{^{'}} + 6 y = 1 + e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

\frac{\partial}{\partial x} ((4 x + 7 y)^{10})

\int x^{2} + x + 2 d x