What is the d/(dy)((y^3)/(39)+(13)/(4y)) ?

The d/(dy)((y^3)/(39)+(13)/(4y)) is (y^2)/(13)-(13)/(4y^2)

What is the first d/(dy)((y^3)/(39)+(13)/(4y)) ?

The first d/(dy)((y^3)/(39)+(13)/(4y)) is (y^2)/(13)-(13)/(4y^2)

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\frac{d}{d y} (\frac{y ^{3}}{39} + \frac{13}{4 y})

\frac{y ^{2}}{13} - \frac{13}{4 y ^{2}}

Solution steps

\frac{d}{d y} (\frac{y ^{3}}{39} + \frac{13}{4 y})

Apply the Sum/Difference Rule:

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

= \frac{d}{d y} (\frac{y ^{3}}{39}) + \frac{d}{d y} (\frac{13}{4 y})

\frac{d}{d y} (\frac{y ^{3}}{39}) = \frac{y ^{2}}{13}

\frac{d}{d y} (\frac{13}{4 y}) = - \frac{13}{4 y ^{2}}

= \frac{y ^{2}}{13} - \frac{13}{4 y ^{2}}

5 x^{4}

\frac{d}{d x} (x^{3} + 3 x y^{2} - 15 x - 12 y)

\frac{d}{d x} (ln (9 x^{2} - 7 x + 1))

\frac{d y}{d t} = 5 - y

\frac{d}{d x} (e^{x} sin (3 x))

What is the d/(dy)((y^3)/(39)+(13)/(4y)) ?
The d/(dy)((y^3)/(39)+(13)/(4y)) is (y^2)/(13)-(13)/(4y^2)
What is the first d/(dy)((y^3)/(39)+(13)/(4y)) ?
The first d/(dy)((y^3)/(39)+(13)/(4y)) is (y^2)/(13)-(13)/(4y^2)