What is the derivative of (60t^2)/(t^2+150) ?

The derivative of (60t^2)/(t^2+150) is (18000t)/((t^2+150)^2)

What is the first derivative of (60t^2)/(t^2+150) ?

The first derivative of (60t^2)/(t^2+150) is (18000t)/((t^2+150)^2)

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

(\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}}

Simplify

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{\partial}{\partial x} (8 x y - \frac{1}{4} (x + y)^{4})

\frac{d}{d x} (cos (\frac{π}{x}))

\int_{2}^{\infty} - 8 x^{- 2} d x

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

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

What is the derivative of (60t^2)/(t^2+150) ?
The derivative of (60t^2)/(t^2+150) is (18000t)/((t^2+150)^2)
What is the first derivative of (60t^2)/(t^2+150) ?
The first derivative of (60t^2)/(t^2+150) is (18000t)/((t^2+150)^2)