What is the derivative of 2/3 (4y^2+1/2)^{3/2} ?

The derivative of 2/3 (4y^2+1/2)^{3/2} is 4*2^{1/2}y(8y^2+1)^{1/2}

What is the first derivative of 2/3 (4y^2+1/2)^{3/2} ?

The first derivative of 2/3 (4y^2+1/2)^{3/2} is 4*2^{1/2}y(8y^2+1)^{1/2}

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derivative of

\frac{2}{3} (4 y^{2} + \frac{1}{2})^{\frac{3}{2}}

4 \cdot 2^{\frac{1}{2}} y (8 y^{2} + 1)^{\frac{1}{2}}

Solution steps

\frac{d}{d y} (\frac{2}{3} (4 y^{2} + \frac{1}{2})^{\frac{3}{2}})

Take the constant out:

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

= \frac{2}{3} \frac{d}{d y} ((4 y^{2} + \frac{1}{2})^{\frac{3}{2}})

Apply the chain rule:

\frac{3 ( 4 y ^{2} + \frac{1}{2} ) ^{\frac{1}{2}}}{2} \frac{d}{d y} (4 y^{2} + \frac{1}{2})

= \frac{3 ( 4 y ^{2} + \frac{1}{2} ) ^{\frac{1}{2}}}{2} \frac{d}{d y} (4 y^{2} + \frac{1}{2})

\frac{d}{d y} (4 y^{2} + \frac{1}{2}) = 8 y

= \frac{2}{3} \cdot \frac{3 ( 4 y ^{2} + \frac{1}{2} ) ^{\frac{1}{2}}}{2} \cdot 8 y

Simplify

\frac{2}{3} \cdot \frac{3 ( 4 y ^{2} + \frac{1}{2} ) ^{\frac{1}{2}}}{2} \cdot 8 y : 4 \cdot 2^{\frac{1}{2}} y (8 y^{2} + 1)^{\frac{1}{2}}

= 4 \cdot 2^{\frac{1}{2}} y (8 y^{2} + 1)^{\frac{1}{2}}

\frac{\partial}{\partial y} (6 x^{2} y)

\frac{d}{d x} (\frac{1}{3} x^{\frac{3}{2}} - x^{5} + π^{4})

\frac{d}{d t} (2 t)

25 y^{^{''}} - y = \frac{x e x}{5}

y^{^{'}} + ln (x) y = ln (x)

What is the derivative of 2/3 (4y^2+1/2)^{3/2} ?
The derivative of 2/3 (4y^2+1/2)^{3/2} is 4*2^{1/2}y(8y^2+1)^{1/2}
What is the first derivative of 2/3 (4y^2+1/2)^{3/2} ?
The first derivative of 2/3 (4y^2+1/2)^{3/2} is 4*2^{1/2}y(8y^2+1)^{1/2}