derivative of arctan((3x-8/(x^3)))

解答

\frac{d}{d x} (arctan (\frac{3 x - 8}{x ^{3}}))

\frac{6 x ^{2} ( - x + 4 )}{( 3 x - 8 ) ^{2} + x ^{6}}

求解步骤

\frac{d}{d x} (arctan (\frac{3 x - 8}{x ^{3}}))

使用连锁律:

\frac{1}{( \frac{3 x - 8}{x ^{3}} ) ^{2} + 1} \frac{d}{d x} (\frac{3 x - 8}{x ^{3}})

= \frac{1}{( \frac{3 x - 8}{x ^{3}} ) ^{2} + 1} \frac{d}{d x} (\frac{3 x - 8}{x ^{3}})

\frac{d}{d x} (\frac{3 x - 8}{x ^{3}}) = \frac{6 ( - x + 4 )}{x ^{4}}

= \frac{1}{( \frac{3 x - 8}{x ^{3}} ) ^{2} + 1} \cdot \frac{6 ( - x + 4 )}{x ^{4}}

化简

\frac{1}{( \frac{3 x - 8}{x ^{3}} ) ^{2} + 1} \cdot \frac{6 ( - x + 4 )}{x ^{4}} : \frac{6 x ^{2} ( - x + 4 )}{( 3 x - 8 ) ^{2} + x ^{6}}

= \frac{6 x ^{2} ( - x + 4 )}{( 3 x - 8 ) ^{2} + x ^{6}}

x \to \infty lim (x^{6})

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

ma c l a u r in x sin (- 2 x), at π

x \to 0 lim (\frac{x}{x ^{2} + 6 x})

\int t + 8 d t