What is the d/(dt)((500)/((t+2))-(1000)/(((t+2)^2))) ?

The d/(dt)((500)/((t+2))-(1000)/(((t+2)^2))) is (-500t+1000)/((t+2)^3)

What is the first d/(dt)((500)/((t+2))-(1000)/(((t+2)^2))) ?

The first d/(dt)((500)/((t+2))-(1000)/(((t+2)^2))) is (-500t+1000)/((t+2)^3)

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\frac{d}{d t} \frac{500}{( t + 2 )} - \frac{1000}{( ( t + 2 ) ^{2} )}

\frac{- 500 t + 1000}{( t + 2 ) ^{3}}

Solution steps

\frac{d}{d t} \frac{500}{( t + 2 )} - \frac{1000}{( ( t + 2 ) ^{2} )}

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (\frac{500}{t + 2}) - \frac{d}{d t} (\frac{1000}{( t + 2 ) ^{2}})

\frac{d}{d t} (\frac{500}{t + 2}) = - \frac{500}{( t + 2 ) ^{2}}

\frac{d}{d t} (\frac{1000}{( t + 2 ) ^{2}}) = - \frac{2000}{( t + 2 ) ^{3}}

= - \frac{500}{( t + 2 ) ^{2}} - (- \frac{2000}{( t + 2 ) ^{3}})

Simplify

- \frac{500}{( t + 2 ) ^{2}} - (- \frac{2000}{( t + 2 ) ^{3}}) : \frac{- 500 t + 1000}{( t + 2 ) ^{3}}

= \frac{- 500 t + 1000}{( t + 2 ) ^{3}}

\frac{x ^{2}}{( e ^{x} - 1 ) ^{2}}

f (x) = \frac{( 0.8 - 0.3 x )}{x}

x^{3} + 4 x^{2} + 3

\frac{d}{d a} (cos^{4} (a) + sin^{4} (a))

y = \frac{x ^{3} - x}{x + 1}

What is the d/(dt)((500)/((t+2))-(1000)/(((t+2)^2))) ?
The d/(dt)((500)/((t+2))-(1000)/(((t+2)^2))) is (-500t+1000)/((t+2)^3)
What is the first d/(dt)((500)/((t+2))-(1000)/(((t+2)^2))) ?
The first d/(dt)((500)/((t+2))-(1000)/(((t+2)^2))) is (-500t+1000)/((t+2)^3)