d/(dy)((y^3)/3+1/(4y))

解答

\frac{d}{d y} (\frac{y ^{3}}{3} + \frac{1}{4 y})

y^{2} - \frac{1}{4 y ^{2}}

求解步骤

\frac{d}{d y} (\frac{y ^{3}}{3} + \frac{1}{4 y})

使用微分加减法定则:

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

= \frac{d}{d y} (\frac{y ^{3}}{3}) + \frac{d}{d y} (\frac{1}{4 y})

\frac{d}{d y} (\frac{y ^{3}}{3}) = y^{2}

\frac{d}{d y} (\frac{1}{4 y}) = - \frac{1}{4 y ^{2}}

= y^{2} - \frac{1}{4 y ^{2}}

\int_{0}^{l n (20)} x e^{x} d x

\frac{\partial}{\partial x} (5 x + 4 y)

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

d er i v a t i v e g (x) = y = 2 x, x = 1

\frac{\partial}{\partial y} (ln (1 + y^{2}))