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

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

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

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

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derivative of

v (t) = (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{7 x ^{6}}{2}

4 e^{5} x

e^{x + 3}

\frac{d}{d r ( x )} (\frac{r ( x ) ^{c o s^{3} (x)}}{( 4 x ^{2} - 1 ) ^{5}})

\frac{d}{d s} (2 \frac{( 590 m )}{9.8} \frac{m}{s ^{2}})

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