derivative of f^{-1}(x)=2x+ln(x+1)

Lösung

\frac{d}{d x} f^{- 1} (x) = 2 x + ln (x + 1)

f (x) = \frac{x}{x ^{2} + x ln ( 1 + x ) + ln ( 1 + x ) - x - 1 + c _{1}}

Schritte zur Lösung

\frac{d}{d x} (f (x)^{- 1} (x)) = 2 x + ln (x + 1)

Wenn

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

dann

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

f (x)^{- 1} x = \int 2 x + ln (x + 1) d x

\int 2 x + ln (x + 1) d x = x^{2} + x ln (x + 1) + ln (x + 1) - x - 1 + c_{1}

f (x)^{- 1} x = x^{2} + x ln (x + 1) + ln (x + 1) - x - 1 + c_{1}

Isoliere

f (x) : f (x) = \frac{x}{x ^{2} + x ln ( 1 + x ) + ln ( 1 + x ) - x - 1 + c _{1}}

f (x) = \frac{x}{x ^{2} + x ln ( 1 + x ) + ln ( 1 + x ) - x - 1 + c _{1}}

\int (1 - x)^{- 2} d x

\frac{\partial ^{2}}{\partial x \partial y} (x y^{2} - 6 x^{2} - 3 y^{2})

s l o p e in t erce pt (7.4) (3.8)

d er i v a t i v e x^{2} + 6 x - 1

\int - 2 x^{4} tan (x^{5}) d x