derivative of 2x^4-6x^3-7x^2+9

解答

\frac{d}{d x} (2 x^{4} - 6 x^{3} - 7 x^{2} + 9)

8 x^{3} - 18 x^{2} - 14 x

求解步骤

\frac{d}{d x} (2 x^{4} - 6 x^{3} - 7 x^{2} + 9)

使用微分加减法定则:

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

= \frac{d}{d x} (2 x^{4}) - \frac{d}{d x} (6 x^{3}) - \frac{d}{d x} (7 x^{2}) + \frac{d}{d x} (9)

\frac{d}{d x} (2 x^{4}) = 8 x^{3}

\frac{d}{d x} (6 x^{3}) = 18 x^{2}

\frac{d}{d x} (7 x^{2}) = 14 x

\frac{d}{d x} (9) = 0

= 8 x^{3} - 18 x^{2} - 14 x + 0

化简

= 8 x^{3} - 18 x^{2} - 14 x

k = 1 \sum \infty k (\frac{2}{3})^{k}

l a pl a ce t r an s f or m sin (t - 1)

\int x^{2} (5 x^{3} + 7 x^{2} - 6 x - 3) d x

sin (2 x) d x + cos (3 y) d y = 0, y (\frac{π}{2}) = \frac{π}{3}

\int \frac{4}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x