What is the first d/(dt)(4t^2+2t+5) ?

The first d/(dt)(4t^2+2t+5) is 8t+2

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d/(dt)(4t^2+2t+5)

\frac{d}{d t} (4 t^{2} + 2 t + 5)

8 t + 2

Solution steps

\frac{d}{d t} (4 t^{2} + 2 t + 5)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (4 t^{2}) + \frac{d}{d t} (2 t) + \frac{d}{d t} (5)

\frac{d}{d t} (4 t^{2}) = 8 t

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

\frac{d}{d t} (5) = 0

= 8 t + 2 + 0

Simplify

= 8 t + 2

\frac{d}{d x} (e^{c o s (x)} + e^{s i n (y)})

\frac{d}{d x} (lo g_{3} (5 x - 1))

y = \frac{5}{x}, (4, \frac{5}{4})

t^{2} \cdot t^{3} + 2

\frac{1}{2 r} + \frac{1}{9 r ^{\frac{8}{9}}}