derivative of-x^5-6x^4-14x^3-16x^2-9x-2

解答

\frac{d}{d x} (- x^{5} - 6 x^{4} - 14 x^{3} - 16 x^{2} - 9 x - 2)

- 5 x^{4} - 24 x^{3} - 42 x^{2} - 32 x - 9

求解步骤

\frac{d}{d x} (- x^{5} - 6 x^{4} - 14 x^{3} - 16 x^{2} - 9 x - 2)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d x} (x^{5}) - \frac{d}{d x} (6 x^{4}) - \frac{d}{d x} (14 x^{3}) - \frac{d}{d x} (16 x^{2}) - \frac{d}{d x} (9 x) - \frac{d}{d x} (2)

\frac{d}{d x} (x^{5}) = 5 x^{4}

\frac{d}{d x} (6 x^{4}) = 24 x^{3}

\frac{d}{d x} (14 x^{3}) = 42 x^{2}

\frac{d}{d x} (16 x^{2}) = 32 x

\frac{d}{d x} (9 x) = 9

\frac{d}{d x} (2) = 0

= - 5 x^{4} - 24 x^{3} - 42 x^{2} - 32 x - 9 - 0

化简

= - 5 x^{4} - 24 x^{3} - 42 x^{2} - 32 x - 9

\frac{\partial ^{2}}{\partial x \partial y} (ln (3 + x^{2} y^{2}))

x \to 0 - lim (\frac{sin ( ∣ x ∣ )}{x})

\int (3 + 2 x) d x

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

\frac{\partial}{\partial s} (\frac{s - r}{t})