What is the derivative of f(t)=sqrt(t/(t^2+7)) ?

The derivative of f(t)=sqrt(t/(t^2+7)) is (-t^2+7)/(2sqrt(t)(t^2+7)^{3/2)}

What is the first derivative of f(t)=sqrt(t/(t^2+7)) ?

The first derivative of f(t)=sqrt(t/(t^2+7)) is (-t^2+7)/(2sqrt(t)(t^2+7)^{3/2)}

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derivative of

f (t) = \frac{t}{t ^{2} + 7}

\frac{- t ^{2} + 7}{2 t ( t ^{2} + 7 ) ^{\frac{3}{2}}}

Solution steps

\frac{d}{d t} (\frac{t}{t ^{2} + 7})

Apply the chain rule:

\frac{1}{2 \frac{t}{t ^{2} + 7}} \frac{d}{d t} (\frac{t}{t ^{2} + 7})

= \frac{1}{2 \frac{t}{t ^{2} + 7}} \frac{d}{d t} (\frac{t}{t ^{2} + 7})

\frac{d}{d t} (\frac{t}{t ^{2} + 7}) = \frac{- t ^{2} + 7}{( t ^{2} + 7 ) ^{2}}

= \frac{1}{2 \frac{t}{t ^{2} + 7}} \cdot \frac{- t ^{2} + 7}{( t ^{2} + 7 ) ^{2}}

Simplify

\frac{1}{2 \frac{t}{t ^{2} + 7}} \cdot \frac{- t ^{2} + 7}{( t ^{2} + 7 ) ^{2}} : \frac{- t ^{2} + 7}{2 t ( t ^{2} + 7 ) ^{\frac{3}{2}}}

= \frac{- t ^{2} + 7}{2 t ( t ^{2} + 7 ) ^{\frac{3}{2}}}

\frac{d}{d x} (e^{x} (6 x^{2} - 12 x + 12))

\frac{d}{d y} (\frac{1}{y ^{2}})

x \to 2 lim (h (x))

\frac{d y}{d x} - 2 x y = - 3 x^{2} e^{x^{2}}

y \to 5 lim (\frac{25 - y ^{2}}{y ^{2} - 4 y - 5})

What is the derivative of f(t)=sqrt(t/(t^2+7)) ?
The derivative of f(t)=sqrt(t/(t^2+7)) is (-t^2+7)/(2sqrt(t)(t^2+7)^{3/2)}
What is the first derivative of f(t)=sqrt(t/(t^2+7)) ?
The first derivative of f(t)=sqrt(t/(t^2+7)) is (-t^2+7)/(2sqrt(t)(t^2+7)^{3/2)}