What is the derivative of sqrt(t/(t^2+8)) ?

The derivative of sqrt(t/(t^2+8)) is (-t^2+8)/(2sqrt(t)(t^2+8)^{3/2)}

What is the first derivative of sqrt(t/(t^2+8)) ?

The first derivative of sqrt(t/(t^2+8)) is (-t^2+8)/(2sqrt(t)(t^2+8)^{3/2)}

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derivative of

\frac{t}{t ^{2} + 8}

\frac{- t ^{2} + 8}{2 t ( t ^{2} + 8 ) ^{\frac{3}{2}}}

Solution steps

\frac{d}{d t} (\frac{t}{t ^{2} + 8})

Apply the chain rule:

\frac{1}{2 \frac{t}{t ^{2} + 8}} \frac{d}{d t} (\frac{t}{t ^{2} + 8})

= \frac{1}{2 \frac{t}{t ^{2} + 8}} \frac{d}{d t} (\frac{t}{t ^{2} + 8})

\frac{d}{d t} (\frac{t}{t ^{2} + 8}) = \frac{- t ^{2} + 8}{( t ^{2} + 8 ) ^{2}}

= \frac{1}{2 \frac{t}{t ^{2} + 8}} \cdot \frac{- t ^{2} + 8}{( t ^{2} + 8 ) ^{2}}

Simplify

\frac{1}{2 \frac{t}{t ^{2} + 8}} \cdot \frac{- t ^{2} + 8}{( t ^{2} + 8 ) ^{2}} : \frac{- t ^{2} + 8}{2 t ( t ^{2} + 8 ) ^{\frac{3}{2}}}

= \frac{- t ^{2} + 8}{2 t ( t ^{2} + 8 ) ^{\frac{3}{2}}}

(2 x - 7) 4 (x^{2} + x + 1) 5

\frac{d}{d x} (- x + 2 cos (2 x))

y = \frac{4 x}{x + 2}, (2, 2)

\int \frac{ln ( 6 x )}{x ^{8}} d x

x \to 1 lim (csc (\frac{x}{2}))

What is the derivative of sqrt(t/(t^2+8)) ?
The derivative of sqrt(t/(t^2+8)) is (-t^2+8)/(2sqrt(t)(t^2+8)^{3/2)}
What is the first derivative of sqrt(t/(t^2+8)) ?
The first derivative of sqrt(t/(t^2+8)) is (-t^2+8)/(2sqrt(t)(t^2+8)^{3/2)}