What is the first d/(dt)(-80-12+2t^2) ?

The first d/(dt)(-80-12+2t^2) is 4t

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d/(dt)(-80-12+2t^2)

\frac{d}{d t} (- 80 - 12 + 2 t^{2})

4 t

Solution steps

\frac{d}{d t} (- 80 - 12 + 2 t^{2})

Simplify

- 80 - 12 + 2 t^{2} : 2 t^{2} - 92

= \frac{d}{d t} (2 t^{2} - 92)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (2 t^{2}) - \frac{d}{d t} (92)

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

\frac{d}{d t} (92) = 0

= 4 t - 0

Simplify

= 4 t

x (1 + x)^{2}

\frac{( 3 x ^{4} + 9 x ^{2} + 1 )}{x}

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

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

(sec (x^{2}))^{3}