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

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

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

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

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

\frac{y ^{3}}{2} - \frac{1}{2 y ^{3}}

Solution steps

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

Simplify

\frac{1}{8} y^{4} + \frac{1}{4 y ^{2}} : \frac{y ^{4}}{8} + \frac{1}{4 y ^{2}}

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

Apply the Sum/Difference Rule:

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

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

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

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

= \frac{y ^{3}}{2} - \frac{1}{2 y ^{3}}

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

y = - 6 x

\int (\frac{2}{x ^{3}} + x^{4}) d x

\frac{d}{d x} (ln (tan (2 x)))

\frac{\partial}{\partial t} (1 - cos (t))

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