## Related questions with answers

Question

Find the derivatives of the functions in the ff exercise.

$v=(1-t)(1+t^2)^{-1}$

Solution

VerifiedStep 1

1 of 2In this exercise we use the following formulas

$\color{#c34632}{\dv{x} (x^n)=nx^{n-1}}$

$\color{#c34632}{\dv{x} (c)=0}$

$\color{#c34632}{\dv{x} (\frac uv)=\frac{v \dv{u}{x}-u\dv{v}{x}}{v^2}}$

The starting equation can be represented as follows

$v=\frac{1-t}{1+t^2}$

Then

$u=1-t \text{ and } s=1+t^2$

Now, we obtain

$\begin{align*} v'&=\dv{v}{t}=\dv{t}(\frac us)\\ &=\frac{s \dv{u}{t}-u\dv{s}{t}}{s^2}\\ &=\frac{(1+t^2)(-1)-(1-t)2t}{(1+t^2)^2}\\ &=\frac{-1-t^2-2t+2t^2}{(1+t^2)^2}\\ &=\frac{t^2-2t-1}{(1+t^2)^2} \end{align*}$

Therefore,

$v'=\frac{t^2-2t-1}{(1+t^2)^2}$

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