(df)/(dx)=4x^2-5x+log_{10}(x)

Lösung

\frac{df}{d x} = 4 x^{2} - 5 x + lo g_{10} (x)

f (x) = \frac{4 x ^{3}}{3} - \frac{5 x ^{2}}{2} + x lo g_{10} (x) - \frac{x}{ln ( 10 )} + c_{1}

Schritte zur Lösung

\frac{df}{d x} = 4 x^{2} - 5 x + lo g_{10} (x)

Ersetze

\frac{df}{d x}

mit

f^{'} (x)

f^{^{'}} (x) = 4 x^{2} - 5 x + lo g_{10} (x)

Wenn

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

dann

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

f (x) = \int 4 x^{2} - 5 x + lo g_{10} (x) d x

\int 4 x^{2} - 5 x + lo g_{10} (x) d x = \frac{4 x ^{3}}{3} - \frac{5 x ^{2}}{2} + x lo g_{10} (x) - \frac{1}{ln ( 10 )} x + c_{1}

f (x) = \frac{4 x ^{3}}{3} - \frac{5 x ^{2}}{2} + x lo g_{10} (x) - \frac{1}{ln ( 10 )} x + c_{1}

Vereinfache

f (x) = \frac{4 x ^{3}}{3} - \frac{5 x ^{2}}{2} + x lo g_{10} (x) - \frac{x}{ln ( 10 )} + c_{1}

\int \frac{2 x}{x - 1} d x

x \to 0 + lim (9)

\frac{\partial}{\partial x} (250 x^{2} + y^{2})

d er i v a t i v e x^{0.8}

d er i v a t i v e \frac{( x ^{2} - 1 )}{x ^{2} - 4}