What is the d/(dy)((y^3)/3+1/(4y)) ?

The d/(dy)((y^3)/3+1/(4y)) is y^2-1/(4y^2)

What is the first d/(dy)((y^3)/3+1/(4y)) ?

The first d/(dy)((y^3)/3+1/(4y)) is y^2-1/(4y^2)

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\frac{d}{d y} (\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}}

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