derivative arctan((1+t)/(1-t))

解答

导数

arctan (\frac{1 + t}{1 - t})

\frac{1}{t ^{2} + 1}

求解步骤

\frac{d}{d t} (arctan (\frac{1 + t}{1 - t}))

使用连锁律:

\frac{1}{( \frac{1 + t}{1 - t} ) ^{2} + 1} \frac{d}{d t} (\frac{1 + t}{1 - t})

= \frac{1}{( \frac{1 + t}{1 - t} ) ^{2} + 1} \frac{d}{d t} (\frac{1 + t}{1 - t})

\frac{d}{d t} (\frac{1 + t}{1 - t}) = \frac{2}{( 1 - t ) ^{2}}

= \frac{1}{( \frac{1 + t}{1 - t} ) ^{2} + 1} \cdot \frac{2}{( 1 - t ) ^{2}}

化简

\frac{1}{( \frac{1 + t}{1 - t} ) ^{2} + 1} \cdot \frac{2}{( 1 - t ) ^{2}} : \frac{1}{t ^{2} + 1}

= \frac{1}{t ^{2} + 1}

\frac{d}{d x} (- 2 (arctan (3 x) - arctan (2 x)) + x)

\int \frac{3}{60 - t} d t

t an g e n t \frac{8}{x ^{2} + 4}

t an g e n t f (x) = \frac{20 x}{x ^{2} - 5}

x \to - 6 + lim (\frac{1}{( x + 6 ) ^{6}})