What is the derivative of ln(t^2+1) ?

The derivative of ln(t^2+1) is (2t)/(t^2+1)

What is the first derivative of ln(t^2+1) ?

The first derivative of ln(t^2+1) is (2t)/(t^2+1)

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derivative of

ln (t^{2} + 1)

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

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d 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}

x \to 0 lim (\frac{x ^{3}}{x})

\int \frac{2 x ^{4} + 7 x ^{3} - x}{x ^{3}} d x

\frac{\partial}{\partial t} (x e^{2 t} - 4 x)

\int_{- 1}^{2} 2^{x} d x

n = 0 \sum \infty e^{- \frac{n}{2}}

What is the derivative of ln(t^2+1) ?
The derivative of ln(t^2+1) is (2t)/(t^2+1)
What is the first derivative of ln(t^2+1) ?
The first derivative of ln(t^2+1) is (2t)/(t^2+1)