What is the derivative of (2-t)(4+t^2)^{-1} ?

The derivative of (2-t)(4+t^2)^{-1} is (t^2-4t-4)/((t^2+4)^2)

What is the first derivative of (2-t)(4+t^2)^{-1} ?

The first derivative of (2-t)(4+t^2)^{-1} is (t^2-4t-4)/((t^2+4)^2)

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derivative of

(2 - t) (4 + t^{2})^{- 1}

\frac{t ^{2} - 4 t - 4}{( t ^{2} + 4 ) ^{2}}

Solution steps

\frac{d}{d t} ((2 - t) (4 + t^{2})^{- 1})

Apply the Product Rule:

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

f = 2 - t, g = (4 + t^{2})^{- 1}

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

\frac{d}{d t} (2 - t) = - 1

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

= (- 1) (4 + t^{2})^{- 1} + (- \frac{2 t}{( 4 + t ^{2} ) ^{2}}) (2 - t)

Simplify

(- 1) (4 + t^{2})^{- 1} + (- \frac{2 t}{( 4 + t ^{2} ) ^{2}}) (2 - t) : \frac{t ^{2} - 4 t - 4}{( t ^{2} + 4 ) ^{2}}

= \frac{t ^{2} - 4 t - 4}{( t ^{2} + 4 ) ^{2}}

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

\int x^{3} y^{4} d x

\frac{d}{d x} (e^{3 x} (- 2 sin (2 x) + 3 cos (2 x)))

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

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

What is the derivative of (2-t)(4+t^2)^{-1} ?
The derivative of (2-t)(4+t^2)^{-1} is (t^2-4t-4)/((t^2+4)^2)
What is the first derivative of (2-t)(4+t^2)^{-1} ?
The first derivative of (2-t)(4+t^2)^{-1} is (t^2-4t-4)/((t^2+4)^2)