What is the d/(dt)(2/(t^2+2)) ?

The d/(dt)(2/(t^2+2)) is -(4t)/((t^2+2)^2)

What is the first d/(dt)(2/(t^2+2)) ?

The first d/(dt)(2/(t^2+2)) is -(4t)/((t^2+2)^2)

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{1}{- u ^{2} - a + b} d u

\frac{d}{d t} (tan (e^{6 t}) + e^{t a n (6 t)})

\frac{\partial}{\partial x} (cos (x y) + e^{yz} + ln (z x))

f^{9}

\frac{d y}{d x} = \frac{( 2 x + 1 )}{3 y}

What is the d/(dt)(2/(t^2+2)) ?
The d/(dt)(2/(t^2+2)) is -(4t)/((t^2+2)^2)
What is the first d/(dt)(2/(t^2+2)) ?
The first d/(dt)(2/(t^2+2)) is -(4t)/((t^2+2)^2)