What is (\partial)/(\partial y)(2y-3/(y^2)+x) ?

The answer to (\partial)/(\partial y)(2y-3/(y^2)+x) is 6/(y^3)+2

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(\partial)/(\partial y)(2y-3/(y^2)+x)

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

\frac{6}{y ^{3}} + 2

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (2 y) = 2

\frac{\partial}{\partial y} (\frac{3}{y ^{2}}) = - \frac{6}{y ^{3}}

\frac{\partial}{\partial y} (x) = 0

= 2 - (- \frac{6}{y ^{3}}) + 0

Simplify

= \frac{6}{y ^{3}} + 2

\int \frac{x ^{4} + 3 x}{x ^{2}} d x

\frac{e ^{x^{2}}}{e ^{x}}

f (x) = e^{x} sin^{2} (x)

\int_{0}^{t} t e^{- t} d t

\frac{\partial}{\partial x} (e^{x - 2})

What is (\partial)/(\partial y)(2y-3/(y^2)+x) ?
The answer to (\partial)/(\partial y)(2y-3/(y^2)+x) is 6/(y^3)+2