(\partial)/(\partial x)([Ax^2+By^3]^{1/4})

Lösung

\frac{\partial}{\partial x} ([A x^{2} + B y^{3}]^{\frac{1}{4}})

\frac{A x}{2 ( A x ^{2} + B y ^{3} ) ^{\frac{3}{4}}}

Schritte zur Lösung

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

Behandle

A, B, y

als Konstante

Wende die Kettenregel an:

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

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

\frac{\partial}{\partial x} (A x^{2} + B y^{3}) = 2 A x

= \frac{1}{4 ( A x ^{2} + B y ^{3} ) ^{\frac{3}{4}}} \cdot 2 A x

Vereinfache

\frac{1}{4 ( A x ^{2} + B y ^{3} ) ^{\frac{3}{4}}} \cdot 2 A x : \frac{A x}{2 ( A x ^{2} + B y ^{3} ) ^{\frac{3}{4}}}

= \frac{A x}{2 ( A x ^{2} + B y ^{3} ) ^{\frac{3}{4}}}

\frac{d}{d x} (\frac{6}{x} - 2)

\int_{0}^{5} π (2 7 y)^{2} d y

\int_{2}^{\infty} e^{- 9 x} d x

\int \frac{1}{4 x y ^{2}} d x

d er i v a t i v e f (x) = \frac{11 x + 7}{3 x ^{4} + 2 x ^{2}}