What is (\partial)/(\partial t)(ln(t^2+1)) ?

The answer to (\partial)/(\partial t)(ln(t^2+1)) is (2t)/(t^2+1)

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(\partial)/(\partial t)(ln(t^2+1))

\frac{\partial}{\partial t} (ln (t^{2} + 1))

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

Solution steps

\frac{\partial}{\partial t} (ln (t^{2} + 1))

Apply the chain rule:

\frac{1}{t ^{2} + 1} \frac{\partial}{\partial t} (t^{2} + 1)

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

\frac{\partial}{\partial t} (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{1}{6 x}

\frac{d}{d x} (\frac{3 x}{x ^{2} + 1})

\int x sin (15 - x) d x

\int \frac{1}{cos ( 2 x )} d x

x \to 0 lim (e^{x l n (x)})

What is (\partial)/(\partial t)(ln(t^2+1)) ?
The answer to (\partial)/(\partial t)(ln(t^2+1)) is (2t)/(t^2+1)