(\partial)/(\partial x)(1(x-y)e^{1y+3x^2})

Lösung

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

6 e^{y + 3 x^{2}} x (x - y) + e^{y + 3 x^{2}}

Schritte zur Lösung

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

Vereinfache

1 \cdot (x - y) e^{1 \cdot y + 3 x^{2}} : e^{y + 3 x^{2}} (x - y)

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

Behandle

y

als Konstante

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{y + 3 x^{2}}, g = x - y

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

\frac{\partial}{\partial x} (e^{y + 3 x^{2}}) = e^{y + 3 x^{2}} \cdot 6 x

\frac{\partial}{\partial x} (x - y) = 1

= e^{y + 3 x^{2}} \cdot 6 x (x - y) + 1 \cdot e^{y + 3 x^{2}}

Vereinfache

= 6 e^{y + 3 x^{2}} x (x - y) + e^{y + 3 x^{2}}

d er i v a t i v e f (x) = \frac{64}{( 1 + 4 x ) ^{3}}

y^{^{'}} = cos^{2} (x) cos^{2} (2 y)

\int \frac{x}{x ^{4} + 49} d x

d er i v a t i v e (x + i y)^{2}

y^{^{'}} + (\frac{2}{x}) y = 0