What is the derivative of (y^3)/9+3/(4y) ?

The derivative of (y^3)/9+3/(4y) is (y^2)/3-3/(4y^2)

What is the first derivative of (y^3)/9+3/(4y) ?

The first derivative of (y^3)/9+3/(4y) is (y^2)/3-3/(4y^2)

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derivative of

\frac{y ^{3}}{9} + \frac{3}{4 y}

\frac{y ^{2}}{3} - \frac{3}{4 y ^{2}}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= \frac{y ^{2}}{3} - \frac{3}{4 y ^{2}}

\frac{1 - 2 x ^{3}}{x ^{4} + 3 x ^{2} - 6}

\int (- 8 y^{3}) d y

\int t^{3} e^{- 5 t} d t

x \to 1 lim (e^{4 x^{3} - 4 x})

What is the derivative of (y^3)/9+3/(4y) ?
The derivative of (y^3)/9+3/(4y) is (y^2)/3-3/(4y^2)
What is the first derivative of (y^3)/9+3/(4y) ?
The first derivative of (y^3)/9+3/(4y) is (y^2)/3-3/(4y^2)