What is (\partial)/(\partial y)(6y+1/x-x/(y^2)) ?

The answer to (\partial)/(\partial y)(6y+1/x-x/(y^2)) is (2x)/(y^3)+6

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(\partial)/(\partial y)(6y+1/x-x/(y^2))

\frac{\partial}{\partial y} (6 y + \frac{1}{x} - \frac{x}{y ^{2}})

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

Solution steps

\frac{\partial}{\partial y} (6 y + \frac{1}{x} - \frac{x}{y ^{2}})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (6 y) + \frac{\partial}{\partial y} (\frac{1}{x}) - \frac{\partial}{\partial y} (\frac{x}{y ^{2}})

\frac{\partial}{\partial y} (6 y) = 6

\frac{\partial}{\partial y} (\frac{1}{x}) = 0

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

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

Simplify

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

\frac{d}{d x} (\frac{x}{x + 8})

\int 3 - x^{2} d x

\frac{1}{s ^{2} + 7 s}

\int 1 + e^{- x} d x

x \to 0 lim (\frac{a}{x})

What is (\partial)/(\partial y)(6y+1/x-x/(y^2)) ?
The answer to (\partial)/(\partial y)(6y+1/x-x/(y^2)) is (2x)/(y^3)+6