What is the first derivative of 3t^2+1 ?

The first derivative of 3t^2+1 is 6t

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derivative of 3t^2+1

derivative of

3 t^{2} + 1

6 t

Solution steps

\frac{d}{d t} (3 t^{2} + 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (3 t^{2}) + \frac{d}{d t} (1)

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

\frac{d}{d t} (1) = 0

= 6 t + 0

Simplify

= 6 t

y = \frac{( 5 x + 9 )}{( x - 1 )}

y = x^{3} + x^{2} + x + 1

\frac{1}{g}

(x - 2)^{3}

tan^{2} (x) - sin^{3} (x)