What is the first (2x^{3/2})^' ?

The first (2x^{3/2})^' is 3x^{1/2}

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(2x^{3/2})^'

(2 x^{\frac{3}{2}})^{^{'}}

3 x^{\frac{1}{2}}

Solution steps

(2 x^{\frac{3}{2}})^{'}

Take the constant out:

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

= 2 (x^{\frac{3}{2}})^{^{'}}

Apply the Power Rule:

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

= 2 \cdot \frac{3}{2} x^{\frac{3}{2} - 1}

Simplify

2 \cdot \frac{3}{2} x^{\frac{3}{2} - 1} : 3 x^{\frac{1}{2}}

= 3 x^{\frac{1}{2}}

t^{2} \frac{d y}{d t} - t = 1 + y + t y, y (1) = 7

\frac{\partial}{\partial x} (x^{2} - y^{3} - x^{2} y + y)

\frac{\partial}{\partial x} (- 2 (1 - x) - 40 x (y - x^{2}))

\frac{d}{d x} (e (- x))

\int_{0.2}^{0.4} 300 x d x