What is the (1/(t-1))^' ?

The (1/(t-1))^' is -1/((t-1)^2)

What is the first (1/(t-1))^' ?

The first (1/(t-1))^' is -1/((t-1)^2)

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(\frac{1}{t - 1})^{^{'}}

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

Solution steps

(\frac{1}{t - 1})^{'}

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= ((t - 1)^{- 1})^{^{'}}

Apply the chain rule:

- \frac{1}{( t - 1 ) ^{2}} (t - 1)^{'}

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

(t - 1)^{'} = 1

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

Simplify

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