derivative of sqrt(y)-xln(y+x)

Lösung

\frac{d}{d x} (y - x ln (y + x))

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

Schritte zur Lösung

\frac{d}{d x} (y - x ln (y + x))

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (y) - \frac{d}{d x} (x ln (y + x))

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

\frac{d}{d x} (x ln (y + x)) = ln (y + x) + \frac{x ( \frac{d y}{d x} + 1 )}{y + x}

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

Vereinfache

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

\int \frac{3}{x ^{2} + 2} - x^{2} d x

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

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

\int \frac{e ^{x}}{e ^{x} + 1 + 1} d x

\frac{\partial}{\partial y} (x^{4} - 2 y^{3})