What is the first d/(dt)(t^2+5t-8) ?

The first d/(dt)(t^2+5t-8) is 2t+5

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d/(dt)(t^2+5t-8)

\frac{d}{d t} (t^{2} + 5 t - 8)

2 t + 5

Solution steps

\frac{d}{d t} (t^{2} + 5 t - 8)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (t^{2}) + \frac{d}{d t} (5 t) - \frac{d}{d t} (8)

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

\frac{d}{d t} (5 t) = 5

\frac{d}{d t} (8) = 0

= 2 t + 5 - 0

Simplify

= 2 t + 5

t^{\frac{1}{4}}

s^{2} + 9

4 x^{3} - 2 x^{2}

\frac{d}{d r} (r^{3} - r^{2})

\frac{d}{d x} (x_{6} x^{2} + 9)