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}}

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