(\partial)/(\partial x)(ln(4x^2+3y^2+6))

Lösung

\frac{\partial}{\partial x} (ln (4 x^{2} + 3 y^{2} + 6))

\frac{8 x}{4 x ^{2} + 3 y ^{2} + 6}

Schritte zur Lösung

\frac{\partial}{\partial x} (ln (4 x^{2} + 3 y^{2} + 6))

Behandle

y

als Konstante

Wende die Kettenregel an:

\frac{1}{4 x ^{2} + 3 y ^{2} + 6} \frac{\partial}{\partial x} (4 x^{2} + 3 y^{2} + 6)

= \frac{1}{4 x ^{2} + 3 y ^{2} + 6} \frac{\partial}{\partial x} (4 x^{2} + 3 y^{2} + 6)

\frac{\partial}{\partial x} (4 x^{2} + 3 y^{2} + 6) = 8 x

= \frac{1}{4 x ^{2} + 3 y ^{2} + 6} \cdot 8 x

Vereinfache

\frac{1}{4 x ^{2} + 3 y ^{2} + 6} \cdot 8 x : \frac{8 x}{4 x ^{2} + 3 y ^{2} + 6}

= \frac{8 x}{4 x ^{2} + 3 y ^{2} + 6}

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

\int e^{5 x} sin (4 x) d x

\frac{d}{d x} ((5 x^{2} - 3) (x^{2} + x + 4))

\frac{d}{d x} (\frac{1}{( 3 x ^{3} + 5 x + 7 ) ^{\frac{3}{4}}})

x \to 16 lim (4 - x)