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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d t}{d t} = 1

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

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

Simplify

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

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

\frac{d}{d x} (\frac{x ^{6}}{3 - x ^{5}})

4 t e^{- t}

x \to - 1 lim (1 + e^{\frac{1}{x}})

\int_{2}^{4} (- x^{2} + 6 x - 8) d x

\frac{d}{d x} (x^{3} (8 x^{2} + 2 x - 3)^{2})