derivative (60t^2)/(t^2+150)

解答

导数

\frac{60 t ^{2}}{t ^{2} + 150}

\frac{18000 t}{( t ^{2} + 150 ) ^{2}}

求解步骤

\frac{d}{d t} (\frac{60 t ^{2}}{t ^{2} + 150})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 60 \frac{d}{d t} (\frac{t ^{2}}{t ^{2} + 150})

使用微分除法定则:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= 60 \cdot \frac{\frac{d}{d t} ( t ^{2} ) ( t ^{2} + 150 ) - \frac{d}{d t} ( t ^{2} + 150 ) t ^{2}}{( t ^{2} + 150 ) ^{2}}

\frac{d}{d t} (t^{2}) = 2 t

\frac{d}{d t} (t^{2} + 150) = 2 t

= 60 \cdot \frac{2 t ( t ^{2} + 150 ) - 2 t t ^{2}}{( t ^{2} + 150 ) ^{2}}

化简

60 \cdot \frac{2 t ( t ^{2} + 150 ) - 2 t t ^{2}}{( t ^{2} + 150 ) ^{2}} : \frac{18000 t}{( t ^{2} + 150 ) ^{2}}

= \frac{18000 t}{( t ^{2} + 150 ) ^{2}}

\frac{d}{d x} (cos (\frac{π}{x}))

x \to 0 lim (\frac{cos ( 3 x )}{x})

\frac{d}{d x} (e^{- x} cos (x))

\frac{d y}{d x} = \frac{y ^{2} + 1}{x ^{2} + 1}

p a r i t y y = cos (tan (x))