What is the derivative of (1-t)(3+t^2)^{-1} ?

The derivative of (1-t)(3+t^2)^{-1} is (t^2-2t-3)/((t^2+3)^2)

What is the first derivative of (1-t)(3+t^2)^{-1} ?

The first derivative of (1-t)(3+t^2)^{-1} is (t^2-2t-3)/((t^2+3)^2)

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derivative of

(1 - t) (3 + t^{2})^{- 1}

\frac{t ^{2} - 2 t - 3}{( t ^{2} + 3 ) ^{2}}

Solution steps

\frac{d}{d t} ((1 - t) (3 + t^{2})^{- 1})

Apply the Product Rule:

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

f = 1 - t, g = (3 + t^{2})^{- 1}

= \frac{d}{d t} (1 - t) (3 + t^{2})^{- 1} + \frac{d}{d t} ((3 + t^{2})^{- 1}) (1 - t)

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

\frac{d}{d t} ((3 + t^{2})^{- 1}) = - \frac{2 t}{( 3 + t ^{2} ) ^{2}}

= (- 1) (3 + t^{2})^{- 1} + (- \frac{2 t}{( 3 + t ^{2} ) ^{2}}) (1 - t)

Simplify

(- 1) (3 + t^{2})^{- 1} + (- \frac{2 t}{( 3 + t ^{2} ) ^{2}}) (1 - t) : \frac{t ^{2} - 2 t - 3}{( t ^{2} + 3 ) ^{2}}

= \frac{t ^{2} - 2 t - 3}{( t ^{2} + 3 ) ^{2}}

y = \frac{sin ( x )}{x}

\frac{x + 5}{x - 5}

y^{^{'}} = 5 x + y

\int x 9 x^{2} + 4 d x

\int x^{2} e^{17 x} d x

What is the derivative of (1-t)(3+t^2)^{-1} ?
The derivative of (1-t)(3+t^2)^{-1} is (t^2-2t-3)/((t^2+3)^2)
What is the first derivative of (1-t)(3+t^2)^{-1} ?
The first derivative of (1-t)(3+t^2)^{-1} is (t^2-2t-3)/((t^2+3)^2)