What is the derivative of ln(t^2+6) ?

The derivative of ln(t^2+6) is (2t)/(t^2+6)

What is the first derivative of ln(t^2+6) ?

The first derivative of ln(t^2+6) is (2t)/(t^2+6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

ln (t^{2} + 6)

\frac{2 t}{t ^{2} + 6}

Solution steps

\frac{d}{d t} (ln (t^{2} + 6))

Apply the chain rule:

\frac{1}{t ^{2} + 6} \frac{d}{d t} (t^{2} + 6)

= \frac{1}{t ^{2} + 6} \frac{d}{d t} (t^{2} + 6)

\frac{d}{d t} (t^{2} + 6) = 2 t

= \frac{1}{t ^{2} + 6} \cdot 2 t

Simplify

\frac{1}{t ^{2} + 6} \cdot 2 t : \frac{2 t}{t ^{2} + 6}

= \frac{2 t}{t ^{2} + 6}

t^{2} e^{t}

\frac{\partial}{\partial x} (sin (4 y) e^{- 5 x})

f (x) = - x^{3} - 150 x^{2} + 60000 x - 800

\int e^{x} (sin (x) + cos (x)) d x

\frac{\partial}{\partial y} (x - 4 y - 9)

What is the derivative of ln(t^2+6) ?
The derivative of ln(t^2+6) is (2t)/(t^2+6)
What is the first derivative of ln(t^2+6) ?
The first derivative of ln(t^2+6) is (2t)/(t^2+6)