(\partial)/(\partial x)(ln(1+x^2)-2e^{5y})

Lösung

\frac{\partial}{\partial x} (ln (1 + x^{2}) - 2 e^{5 y})

\frac{2 x}{x ^{2} + 1}

Schritte zur Lösung

\frac{\partial}{\partial x} (ln (1 + x^{2}) - 2 e^{5 y})

Behandle

y

als Konstante

Wende die Summen-/Differenzregel an:

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

= \frac{\partial}{\partial x} (ln (1 + x^{2})) - \frac{\partial}{\partial x} (2 e^{5 y})

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

\frac{\partial}{\partial x} (2 e^{5 y}) = 0

= \frac{2 x}{1 + x ^{2}} - 0

Vereinfache

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

\int \frac{1}{x ( ln ( x ) ) ^{p}} d x

\frac{d}{d y} (c y^{- 6})

x \to 0 lim (cos (x) - sin (x))

\frac{d}{d x} (\frac{e ^{x}}{5 x})

\int - 3 arcsin (10 x) d x