derivative y=(5+xf(x))/(sqrt(x))

Lösung

ableitung von

y = \frac{5 + x f ( x )}{x}

\frac{2 x ^{2} \frac{df}{d x} + x f ( x ) - 5}{2 x x}

Schritte zur Lösung

\frac{d}{d x} (\frac{5 + x f ( x )}{x})

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (5 + x f (x)) = x \frac{df}{d x} + f (x)

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

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

Vereinfache

\frac{( x \frac{df}{d x} + f ( x ) ) x - \frac{1}{2 x} ( 5 + x f ( x ) )}{( x ) ^{2}} : \frac{2 x ^{2} \frac{df}{d x} + x f ( x ) - 5}{2 x x}

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

\int 9 x^{4} d x

y^{^{'}} + 2 x y = 2 x e^{- x^{2}}

\frac{d y}{d x} = \frac{y ^{2} x}{x ^{2} + 1}

y^{^{''}} - 8 y^{^{'}} + 16 y = \frac{12 e ^{4 t}}{( t ^{2} + 1 )}

t an g e n t y = sin (x) cos (x), at x = \frac{3 π}{4}