What is the d/(dt)(-1/(2(t+1) 3/2)) ?

The d/(dt)(-1/(2(t+1) 3/2)) is 1/(3(t+1)^2)

What is the first d/(dt)(-1/(2(t+1) 3/2)) ?

The first d/(dt)(-1/(2(t+1) 3/2)) is 1/(3(t+1)^2)

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\frac{d}{d t} (- \frac{1}{2 ( t + 1 ) \frac{3}{2}})

\frac{1}{3 ( t + 1 ) ^{2}}

Solution steps

\frac{d}{d t} (- \frac{1}{2 ( t + 1 ) \frac{3}{2}})

Simplify

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

= \frac{d}{d t} (- \frac{1}{3 ( t + 1 )})

Take the constant out:

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

= - \frac{1}{3} \frac{d}{d t} (\frac{1}{t + 1})

Apply exponent rule:

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

= - \frac{1}{3} \frac{d}{d t} ((t + 1)^{- 1})

Apply the chain rule:

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

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

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

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

Simplify

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

= \frac{1}{3 ( t + 1 ) ^{2}}

x \to - 2 + lim (\frac{3 x - 6}{x + 2})

\frac{d}{d x} (x^{3} sin (x) cos (x))

(x^{5} + y^{2}) d x - x y d y = 0

x \to 7 lim (6 x + ∣ x - 7 ∣)

x \to - 5 lim (x^{2} - 10)

What is the d/(dt)(-1/(2(t+1) 3/2)) ?
The d/(dt)(-1/(2(t+1) 3/2)) is 1/(3(t+1)^2)
What is the first d/(dt)(-1/(2(t+1) 3/2)) ?
The first d/(dt)(-1/(2(t+1) 3/2)) is 1/(3(t+1)^2)