What is the d/(dy)((y^3)/(27)+9/(4y)) ?

The d/(dy)((y^3)/(27)+9/(4y)) is (y^2)/9-9/(4y^2)

What is the first d/(dy)((y^3)/(27)+9/(4y)) ?

The first d/(dy)((y^3)/(27)+9/(4y)) is (y^2)/9-9/(4y^2)

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\frac{d}{d y} (\frac{y ^{3}}{27} + \frac{9}{4 y})

\frac{y ^{2}}{9} - \frac{9}{4 y ^{2}}

Solution steps

\frac{d}{d y} (\frac{y ^{3}}{27} + \frac{9}{4 y})

Apply the Sum/Difference Rule:

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

= \frac{d}{d y} (\frac{y ^{3}}{27}) + \frac{d}{d y} (\frac{9}{4 y})

\frac{d}{d y} (\frac{y ^{3}}{27}) = \frac{y ^{2}}{9}

\frac{d}{d y} (\frac{9}{4 y}) = - \frac{9}{4 y ^{2}}

= \frac{y ^{2}}{9} - \frac{9}{4 y ^{2}}

sin (4 x) + sin^{2} (4 x)

f (x) = - 3 lo g_{10} (x)

f (x) = - 2 x

f^{^{'}} (x) = 5 x^{2} + 7

4 sec (x) - 8 cos (x)

What is the d/(dy)((y^3)/(27)+9/(4y)) ?
The d/(dy)((y^3)/(27)+9/(4y)) is (y^2)/9-9/(4y^2)
What is the first d/(dy)((y^3)/(27)+9/(4y)) ?
The first d/(dy)((y^3)/(27)+9/(4y)) is (y^2)/9-9/(4y^2)