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)}

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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}}}

\int e^{x} - 2 d x

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\int_{0}^{l n (3)} \frac{3}{e ^{x} + 4} d x

\frac{d y}{d x} + 2 y = e^{x}

\int \frac{1}{( 1 - x ) ( 5 - x )} d x

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)}