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

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

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d/(dt)(1/2 t^2+t)

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

t + 1

Solution steps

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

Simplify

\frac{1}{2} t^{2} + t : \frac{t ^{2}}{2} + t

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d t}{d t} = 1

= t + 1

f (x) = 3^{- 6 x + 6}

\int_{0}^{2} \frac{2 x}{( x ^{2} + 1 ) ^{3}} d x

\frac{d}{d x} (3 x^{2} - 6 x + 1)

\frac{1}{( s - 2 ) ^{2} - 4}

y^{^{''''}} - 2 y^{^{''}} + y = 0