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

The derivative of ((y^3)/3)+1/(4y) is y^2-1/(4y^2)

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

The first derivative of ((y^3)/3)+1/(4y) is y^2-1/(4y^2)

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derivative of

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

y^{2} - \frac{1}{4 y ^{2}}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

f (x) = \frac{x \cdot , ( x )}{1 + x}

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

f (x) = (3 x + 4) (5 - x^{2}) (4 + x^{2})

x \to \infty lim (e^{x} + 3)

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