derivative of 2x+y^3=x

Lösung

\frac{d}{d x} (2 x + y)^{3} = x

y = 3 \frac{x ^{2} + c _{1}}{2} - 2 x, y = \frac{2 ^{\frac{2}{3}} ( - 3 x ^{2} + 2 c _{1} - 4 3 2 x )}{4} + i \frac{2 ^{\frac{2}{3}} 3 3 x ^{2} + 2 c _{1}}{4}, y = \frac{2 ^{\frac{2}{3}} ( - 3 x ^{2} + 2 c _{1} - 4 3 2 x )}{4} - i \frac{2 ^{\frac{2}{3}} 3 3 x ^{2} + 2 c _{1}}{4}

Schritte zur Lösung

\frac{d}{d x} ((2 x + y)^{3}) = x

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

(2 x + y)^{3} = \int x d x

\int x d x = \frac{x ^{2}}{2} + c_{1}

(2 x + y)^{3} = \frac{x ^{2}}{2} + c_{1}

Isoliere

y : y = 3 \frac{x ^{2} + 2 c _{1}}{2} - 2 x, y = \frac{2 ^{\frac{2}{3}} ( - 3 x ^{2} + 2 c _{1} - 4 3 2 x )}{4} + i \frac{2 ^{\frac{2}{3}} 3 3 x ^{2} + 2 c _{1}}{4}, y = \frac{2 ^{\frac{2}{3}} ( - 3 x ^{2} + 2 c _{1} - 4 3 2 x )}{4} - i \frac{2 ^{\frac{2}{3}} 3 3 x ^{2} + 2 c _{1}}{4}

y = 3 \frac{x ^{2} + 2 c _{1}}{2} - 2 x, y = \frac{2 ^{\frac{2}{3}} ( - 3 x ^{2} + 2 c _{1} - 4 3 2 x )}{4} + i \frac{2 ^{\frac{2}{3}} 3 3 x ^{2} + 2 c _{1}}{4}, y = \frac{2 ^{\frac{2}{3}} ( - 3 x ^{2} + 2 c _{1} - 4 3 2 x )}{4} - i \frac{2 ^{\frac{2}{3}} 3 3 x ^{2} + 2 c _{1}}{4}

Vereinfache

y = 3 \frac{x ^{2} + c _{1}}{2} - 2 x, y = \frac{2 ^{\frac{2}{3}} ( - 3 x ^{2} + 2 c _{1} - 4 3 2 x )}{4} + i \frac{2 ^{\frac{2}{3}} 3 3 x ^{2} + 2 c _{1}}{4}, y = \frac{2 ^{\frac{2}{3}} ( - 3 x ^{2} + 2 c _{1} - 4 3 2 x )}{4} - i \frac{2 ^{\frac{2}{3}} 3 3 x ^{2} + 2 c _{1}}{4}

\int \frac{x + 16}{x} d x

x \to - 3 lim (- (3 x))

d er i v a t i v e e^{(3 h^{2} + 6)}

t \to 0 lim (\frac{( e ^{t} - 1 )}{t})

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