What is the d/(dt)((t+1)/(t-1)) ?

The d/(dt)((t+1)/(t-1)) is -2/((t-1)^2)

What is the first d/(dt)((t+1)/(t-1)) ?

The first d/(dt)((t+1)/(t-1)) is -2/((t-1)^2)

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

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

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

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

x, x^{\frac{1}{2}}, 1, 2

f (x) = 3 sin (x), at x = \frac{π}{3}

f (x) = 2^{x} ln (2)

y^{^{'}} = y + x^{2}

What is the d/(dt)((t+1)/(t-1)) ?
The d/(dt)((t+1)/(t-1)) is -2/((t-1)^2)
What is the first d/(dt)((t+1)/(t-1)) ?
The first d/(dt)((t+1)/(t-1)) is -2/((t-1)^2)