What is the first derivative of 4t^2+1 ?

The first derivative of 4t^2+1 is 8t

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derivative of 4t^2+1

derivative of

4 t^{2} + 1

8 t

Solution steps

\frac{d}{d t} (4 t^{2} + 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (4 t^{2}) + \frac{d}{d t} (1)

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

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

= 8 t + 0

Simplify

= 8 t

(1 + x^{2}) + (1 + y) y^{^{'}} = 0

y = \frac{4}{3} π x^{3}

e^{\frac{z}{( z - 1 )}}

\frac{d}{d x} ((1 - x^{\frac{2}{3}})^{\frac{3}{2}})

x^{2} + 7 y^{2} = 7