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Question

Find the derivatives of the functions in the ff exercise.

v=(1t)(1+t2)1v=(1-t)(1+t^2)^{-1}

Solution

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In this exercise we use the following formulas

d ⁣dx(xn)=nxn1\color{#c34632}{\dv{x} (x^n)=nx^{n-1}}

d ⁣dx(c)=0\color{#c34632}{\dv{x} (c)=0}

d ⁣dx(uv)=v ⁣du ⁣dxu ⁣dv ⁣dxv2\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=1t1+t2v=\frac{1-t}{1+t^2}

Then

u=1t and s=1+t2u=1-t \text{ and } s=1+t^2

Now, we obtain

v= ⁣dv ⁣dt=d ⁣dt(us)=s ⁣du ⁣dtu ⁣ds ⁣dts2=(1+t2)(1)(1t)2t(1+t2)2=1t22t+2t2(1+t2)2=t22t1(1+t2)2\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=t22t1(1+t2)2v'=\frac{t^2-2t-1}{(1+t^2)^2}

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