What is (\partial)/(\partial x)(-y/(x^2)+4x+1/y) ?

The answer to (\partial)/(\partial x)(-y/(x^2)+4x+1/y) is (2y)/(x^3)+4

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(\partial)/(\partial x)(-y/(x^2)+4x+1/y)

\frac{\partial}{\partial x} (- \frac{y}{x ^{2}} + 4 x + \frac{1}{y})

\frac{2 y}{x ^{3}} + 4

Solution steps

\frac{\partial}{\partial x} (- \frac{y}{x ^{2}} + 4 x + \frac{1}{y})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= - \frac{\partial}{\partial x} (\frac{y}{x ^{2}}) + \frac{\partial}{\partial x} (4 x) + \frac{\partial}{\partial x} (\frac{1}{y})

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

\frac{\partial}{\partial x} (4 x) = 4

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

= - (- \frac{2 y}{x ^{3}}) + 4 + 0

Simplify

= \frac{2 y}{x ^{3}} + 4

x^{2} \frac{d y}{d x} - 2 x y = 5 y^{4}

\int_{- 2}^{2} (x + 2) 4 - x^{2} d x

y^{^{''}} + 12 y^{^{'}} + 40 y = 0

\int - 2 e^{- 2 x} d x

y^{^{''}} + 16 y = - 8 sin (4 t)

What is (\partial)/(\partial x)(-y/(x^2)+4x+1/y) ?
The answer to (\partial)/(\partial x)(-y/(x^2)+4x+1/y) is (2y)/(x^3)+4