What is the d/(dt)(65+2t-3t^2) ?

The d/(dt)(65+2t-3t^2) is -6t+2

What is the first d/(dt)(65+2t-3t^2) ?

The first d/(dt)(65+2t-3t^2) is -6t+2

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\frac{d}{d t} (65 + 2 t - 3 t^{2})

- 6 t + 2

Solution steps

\frac{d}{d t} (65 + 2 t - 3 t^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (65) + \frac{d}{d t} (2 t) - \frac{d}{d t} (3 t^{2})

\frac{d}{d t} (65) = 0

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

\frac{d}{d t} (3 t^{2}) = 6 t

= 0 + 2 - 6 t

Simplify

= - 6 t + 2

- 2 ln (x)

f (x) = (x + sin (x))^{2}

\frac{d}{d x} (ln ((x - b)^{2}) - ln ((y - a)^{2}))

f (x) = x^{2} - 4 x + 9

e^{x} - y^{2}