What is the first derivative of 2y^2 ?

The first derivative of 2y^2 is 4y

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derivative of 2y^2

derivative of

2 y^{2}

4 y

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 \cdot 2 y^{2 - 1}

Simplify

= 4 y

\frac{d}{d x} (cos^{3} (sin (2 x)))

15 x^{4} - 4 x^{3}

\int 10 x^{\frac{3}{7}} + 7 x^{- \frac{6}{7}} d x

x \to 10 lim (x^{2})

y^{^{''}} + 0.2121 y^{^{'}} + 2 y = cos (4 x), y (0) = 0