What is the d/(dy)(3x^2y^2+x/y) ?

The d/(dy)(3x^2y^2+x/y) is 6x^2y-x/(y^2)

What is the first d/(dy)(3x^2y^2+x/y) ?

The first d/(dy)(3x^2y^2+x/y) is 6x^2y-x/(y^2)

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

6 x^{2} y - \frac{x}{y ^{2}}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{d}{d y} (3 x^{2} y^{2}) = 6 x^{2} y

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

= 6 x^{2} y - \frac{x}{y ^{2}}

\frac{d}{d x} ((3 x^{4} - 2) (1 + 4 x^{3}))

- \frac{14}{x ^{3}}

4 x^{5} + 11

\frac{\partial}{\partial x} (- 8 d (J + \frac{1}{2})^{2})

y^{^{'}} - 5 y = e^{- 2 x} y^{- 2}

What is the d/(dy)(3x^2y^2+x/y) ?
The d/(dy)(3x^2y^2+x/y) is 6x^2y-x/(y^2)
What is the first d/(dy)(3x^2y^2+x/y) ?
The first d/(dy)(3x^2y^2+x/y) is 6x^2y-x/(y^2)