What is the first derivative of 4y^2 ?

The first derivative of 4y^2 is 8y

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derivative of 4y^2

derivative of

4 y^{2}

8 y

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule:

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

= 4 \cdot 2 y^{2 - 1}

Simplify

= 8 y

t an g e n t y = \frac{6 x}{x + 3}, (- 1, - 3)

d er i v a t i v e f (x) = \frac{2 x}{x ^{2} - 8}

\int \frac{1}{t ^{2} - 2 t + 5} d t

x \to - 1 + lim (x + 12)

x \to 2 lim (\frac{x}{x - 1})