derivative of ln((x^2+1/(x+4)))

解答

\frac{d}{d x} (ln (\frac{x ^{2} + 1}{x + 4}))

\frac{x ^{2} + 8 x - 1}{( x ^{2} + 1 ) ( x + 4 )}

求解步骤

\frac{d}{d x} (ln (\frac{x ^{2} + 1}{x + 4}))

使用连锁律:

\frac{1}{\frac{x ^{2} + 1}{x + 4}} \frac{d}{d x} (\frac{x ^{2} + 1}{x + 4})

= \frac{1}{\frac{x ^{2} + 1}{x + 4}} \frac{d}{d x} (\frac{x ^{2} + 1}{x + 4})

\frac{d}{d x} (\frac{x ^{2} + 1}{x + 4}) = \frac{x ^{2} + 8 x - 1}{( x + 4 ) ^{2}}

= \frac{1}{\frac{x ^{2} + 1}{x + 4}} \cdot \frac{x ^{2} + 8 x - 1}{( x + 4 ) ^{2}}

化简

\frac{1}{\frac{x ^{2} + 1}{x + 4}} \cdot \frac{x ^{2} + 8 x - 1}{( x + 4 ) ^{2}} : \frac{x ^{2} + 8 x - 1}{( x ^{2} + 1 ) ( x + 4 )}

= \frac{x ^{2} + 8 x - 1}{( x ^{2} + 1 ) ( x + 4 )}

\int e^{7 x + 9} d x

\int \frac{sin ( x )}{2 - cos ^{2} ( x )} d x

d er i v a t i v e e^{4 x} cos (7 x)

\frac{d}{d x} (120 x)

x \to 0 lim (\frac{4 x ^{2} - 3 x}{x ^{2} + 0.8 x + \frac{4}{x}})