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

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

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

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

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

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

Solution steps

\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}}

f (x) = \frac{36}{( 1 + 3 x ) ^{3}}

\int_{0}^{6 π} t^{2} sin (2 t) d t

\int 6 (x - 12)^{5} d x

f (x) = x^{2} - 6, at x = 3

6 e^{z}

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