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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial y} (ln (x - 9 y))

a re a y = ∣ x - 9 ∣, y = \frac{x}{2}

l a pl a ce t r an s f or m x^{2} cos (x)

d er i v a t i v e f (x) = (5 x^{6} + 4 x^{3})^{4}

(4 - t^{2}) y^{^{'}} + 2 t y = 3 t^{2}

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