d/(dt)(arctan(4t))

解答

\frac{d}{d t} (arctan (4 t))

\frac{4}{16 t ^{2} + 1}

求解步骤

\frac{d}{d t} (arctan (4 t))

使用连锁律:

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

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

\frac{d}{d t} (4 t) = 4

= \frac{1}{( 4 t ) ^{2} + 1} \cdot 4

化简

\frac{1}{( 4 t ) ^{2} + 1} \cdot 4 : \frac{4}{16 t ^{2} + 1}

= \frac{4}{16 t ^{2} + 1}

x \to 0 + lim (x ln (tan (5 x)))

in v erse l a pl a ce 4 s

t an g e n t f (x) = \frac{2}{x}, at x = - 9

\frac{d}{d x} ((x^{2} + 2 x) e^{x})

t an g e n t x^{2} + 2 x y - y^{2} + x = 51, (5, 7)