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

The d/(dy)((y/8)^4+1/(4y^2)) is (y^3)/(1024)-1/(2y^3)

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

The first d/(dy)((y/8)^4+1/(4y^2)) is (y^3)/(1024)-1/(2y^3)

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

\frac{y ^{3}}{1024} - \frac{1}{2 y ^{3}}

Solution steps

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

Simplify

(\frac{y}{8})^{4} + \frac{1}{4 y ^{2}} : \frac{y ^{4}}{4096} + \frac{1}{4 y ^{2}}

= \frac{d}{d y} (\frac{y ^{4}}{4096} + \frac{1}{4 y ^{2}})

Apply the Sum/Difference Rule:

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

= \frac{d}{d y} (\frac{y ^{4}}{4096}) + \frac{d}{d y} (\frac{1}{4 y ^{2}})

\frac{d}{d y} (\frac{y ^{4}}{4096}) = \frac{y ^{3}}{1024}

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

= \frac{y ^{3}}{1024} - \frac{1}{2 y ^{3}}

x^{2} y^{^{''}} - 2 y = 0

x \to 1 lim (cos (π) x)

n = 1 \sum \infty 4 (\frac{2}{3})^{n - 1}

\int \frac{1}{x ^{3} + x} d x

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

What is the d/(dy)((y/8)^4+1/(4y^2)) ?
The d/(dy)((y/8)^4+1/(4y^2)) is (y^3)/(1024)-1/(2y^3)
What is the first d/(dy)((y/8)^4+1/(4y^2)) ?
The first d/(dy)((y/8)^4+1/(4y^2)) is (y^3)/(1024)-1/(2y^3)