What is the d/(dt)(ln(t^2+1)) ?

The d/(dt)(ln(t^2+1)) is (2t)/(t^2+1)

What is the first d/(dt)(ln(t^2+1)) ?

The first d/(dt)(ln(t^2+1)) is (2t)/(t^2+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d t} (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}

y^{^{'}} = y (1 - y), y (0) = 6

y^{^{'}} = 1 - y^{2} sin (x)

x \to 0 + lim (\frac{x e ^{\frac{1}{x}}}{2})

\int 9 x sin^{2} (x) d x

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

What is the d/(dt)(ln(t^2+1)) ?
The d/(dt)(ln(t^2+1)) is (2t)/(t^2+1)
What is the first d/(dt)(ln(t^2+1)) ?
The first d/(dt)(ln(t^2+1)) is (2t)/(t^2+1)