What is the (ln(t^2+1))^' ?

The (ln(t^2+1))^' is (2t)/(t^2+1)

What is the first (ln(t^2+1))^' ?

The first (ln(t^2+1))^' is (2t)/(t^2+1)

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(ln (t^{2} + 1))^{^{'}}

\frac{2 t}{t ^{2} + 1}

Solution steps

(ln (t^{2} + 1))^{'}

Apply the chain rule:

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

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

(t^{2} + 1)^{'} = 2 t

= \frac{1}{t ^{2} + 1} \cdot 2 t

Simplify

\frac{1}{t ^{2} + 1} \cdot 2 t : \frac{2 t}{t ^{2} + 1}

= \frac{2 t}{t ^{2} + 1}

\frac{d y}{d t} = y - t

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