derivative of (2x^3-3log_{2}(5-x))

解答

\frac{d}{d x} ((2 x^{3} - 3) lo g_{2} (5 - x))

\frac{- 2 x ^{3} + 3 + 30 x ^{2} ln ( 5 - x ) - 6 x ^{3} ln ( 5 - x )}{ln ( 2 ) ( 5 - x )}

求解步骤

\frac{d}{d x} ((2 x^{3} - 3) lo g_{2} (5 - x))

使用微分乘法定则:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 2 x^{3} - 3, g = lo g_{2} (5 - x)

= \frac{d}{d x} (2 x^{3} - 3) lo g_{2} (5 - x) + \frac{d}{d x} (lo g_{2} (5 - x)) (2 x^{3} - 3)

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

\frac{d}{d x} (lo g_{2} (5 - x)) = - \frac{1}{ln ( 2 ) ( 5 - x )}

= 6 x^{2} lo g_{2} (5 - x) + (- \frac{1}{ln ( 2 ) ( 5 - x )}) (2 x^{3} - 3)

化简

6 x^{2} lo g_{2} (5 - x) + (- \frac{1}{ln ( 2 ) ( 5 - x )}) (2 x^{3} - 3) : \frac{- 2 x ^{3} + 3 + 30 x ^{2} ln ( 5 - x ) - 6 x ^{3} ln ( 5 - x )}{ln ( 2 ) ( 5 - x )}

= \frac{- 2 x ^{3} + 3 + 30 x ^{2} ln ( 5 - x ) - 6 x ^{3} ln ( 5 - x )}{ln ( 2 ) ( 5 - x )}

t an g e n t f (x) = \frac{1}{4 + 3 x}, at x = 1

\frac{\partial}{\partial v} (\frac{u + v}{3})

x \cdot \frac{d y}{d x} - (1 + x) y = x y^{2}

a re a x^{2}, 0, 1

n = 0 \sum \infty (- 1)^{n} x (\frac{x}{4})^{2 n}