derivative of (sqrt(x)+1/(sqrt(x)+3))

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

\frac{\partial}{\partial x} (\frac{2}{3} y^{- \frac{1}{3}})

y^{^{'}} = (x + y - 1)^{2}, y (0) = 0

\frac{d}{d x} (\frac{4 ( - x ^{2} - 4 )}{( x ^{2} - 4 ) ^{2}})

(x^{2} sin (x^{2}))^{^{'}}

\int \frac{( 4 x ^{3} - 6 x ^{4} )}{2 x ^{6}} d x