What is the first d/(dt)(5t+16t^2) ?

The first d/(dt)(5t+16t^2) is 5+32t

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d/(dt)(5t+16t^2)

\frac{d}{d t} (5 t + 16 t^{2})

5 + 32 t

Solution steps

\frac{d}{d t} (5 t + 16 t^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (5 t) + \frac{d}{d t} (16 t^{2})

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

\frac{d}{d t} (16 t^{2}) = 32 t

= 5 + 32 t

x (1 + x)^{2}

e^{x^{2} - y^{2}}

(8 x^{2} + 3 x + 1) \cdot 0.2 x

- 2 + 3 sin (x)

\frac{d}{d p} (20 p)