What is the first derivative of 2y^2+1 ?

The first derivative of 2y^2+1 is 4y

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derivative of 2y^2+1

derivative of

2 y^{2} + 1

4 y

Solution steps

\frac{d}{d y} (2 y^{2} + 1)

Apply the Sum/Difference Rule:

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

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

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

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

= 4 y + 0

Simplify

= 4 y

\frac{d}{d t} (a (cos (t) + lo g_{10} (\frac{tan ( t )}{2})))

d er i v a t i v e \frac{( 1 + x )}{( 1 + x ^{2} )}

d er i v a t i v e \frac{x + 1}{2 x - 1}

d er i v a t i v e sin (y^{2})

d er i v a t i v e y = sin^{2} (x) 0.2 x