What is the d/(dy)((y^4)/4+1/(8y^2)) ?

The d/(dy)((y^4)/4+1/(8y^2)) is y^3-1/(4y^3)

What is the first d/(dy)((y^4)/4+1/(8y^2)) ?

The first d/(dy)((y^4)/4+1/(8y^2)) is y^3-1/(4y^3)

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

y^{3} - \frac{1}{4 y ^{3}}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d y} (\frac{1}{8 y ^{2}}) = - \frac{1}{4 y ^{3}}

= y^{3} - \frac{1}{4 y ^{3}}

\frac{d ^{2}}{d x ^{2}} (x x^{2} + 1)

\int 2 x^{\frac{1}{4}} - 3 x^{3} d x

\frac{d}{d θ} (14 sin (θ))

\frac{d y}{d x} = 2 x e^{- y}

\int \frac{1}{a - b y} d y

What is the d/(dy)((y^4)/4+1/(8y^2)) ?
The d/(dy)((y^4)/4+1/(8y^2)) is y^3-1/(4y^3)
What is the first d/(dy)((y^4)/4+1/(8y^2)) ?
The first d/(dy)((y^4)/4+1/(8y^2)) is y^3-1/(4y^3)