What is the first d/(dt)(2t^2-10) ?

The first d/(dt)(2t^2-10) is 4t

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d/(dt)(2t^2-10)

\frac{d}{d t} (2 t^{2} - 10)

4 t

Solution steps

\frac{d}{d t} (2 t^{2} - 10)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (2 t^{2}) - \frac{d}{d t} (10)

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

\frac{d}{d t} (10) = 0

= 4 t - 0

Simplify

= 4 t

\int - \frac{5}{3 x} d x

\frac{( 2 x ^{4} + 5 x ^{3} - 2 x - 8 )}{( x + 3 )}

\frac{\partial}{\partial x} (\frac{4 x - 3 t}{s + 2 x})

\frac{\partial}{\partial x} (x - cos (x))

y^{^{'}} = y x