What is the derivative of sqrt(t/(t^2+4)) ?

The derivative of sqrt(t/(t^2+4)) is (-t^2+4)/(2sqrt(t)(t^2+4)^{3/2)}

What is the first derivative of sqrt(t/(t^2+4)) ?

The first derivative of sqrt(t/(t^2+4)) is (-t^2+4)/(2sqrt(t)(t^2+4)^{3/2)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

derivative of

\frac{t}{t ^{2} + 4}

\frac{- t ^{2} + 4}{2 t ( t ^{2} + 4 ) ^{\frac{3}{2}}}

Solution steps

\frac{d}{d t} (\frac{t}{t ^{2} + 4})

Apply the chain rule:

\frac{1}{2 \frac{t}{t ^{2} + 4}} \frac{d}{d t} (\frac{t}{t ^{2} + 4})

= \frac{1}{2 \frac{t}{t ^{2} + 4}} \frac{d}{d t} (\frac{t}{t ^{2} + 4})

\frac{d}{d t} (\frac{t}{t ^{2} + 4}) = \frac{- t ^{2} + 4}{( t ^{2} + 4 ) ^{2}}

= \frac{1}{2 \frac{t}{t ^{2} + 4}} \cdot \frac{- t ^{2} + 4}{( t ^{2} + 4 ) ^{2}}

Simplify

\frac{1}{2 \frac{t}{t ^{2} + 4}} \cdot \frac{- t ^{2} + 4}{( t ^{2} + 4 ) ^{2}} : \frac{- t ^{2} + 4}{2 t ( t ^{2} + 4 ) ^{\frac{3}{2}}}

= \frac{- t ^{2} + 4}{2 t ( t ^{2} + 4 ) ^{\frac{3}{2}}}

What is the derivative of sqrt(t/(t^2+4)) ?
The derivative of sqrt(t/(t^2+4)) is (-t^2+4)/(2sqrt(t)(t^2+4)^{3/2)}
What is the first derivative of sqrt(t/(t^2+4)) ?
The first derivative of sqrt(t/(t^2+4)) is (-t^2+4)/(2sqrt(t)(t^2+4)^{3/2)}