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

The answer to (\partial)/(\partial x)(x^2 y/(x^4+y^2)) is (y(-2x^5+2xy^2))/((x^4+y^2)^2)

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

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

\frac{y ( - 2 x ^{5} + 2 x y ^{2} )}{( x ^{4} + y ^{2} ) ^{2}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2}, g = \frac{1}{x ^{4} + y ^{2}}

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

\frac{\partial}{\partial x} (x^{2}) = 2 x

\frac{\partial}{\partial x} (\frac{1}{x ^{4} + y ^{2}}) = - \frac{4 x ^{3}}{( x ^{4} + y ^{2} ) ^{2}}

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

Simplify

y (2 x \frac{1}{x ^{4} + y ^{2}} + (- \frac{4 x ^{3}}{( x ^{4} + y ^{2} ) ^{2}}) x^{2}) : \frac{y ( - 2 x ^{5} + 2 x y ^{2} )}{( x ^{4} + y ^{2} ) ^{2}}

= \frac{y ( - 2 x ^{5} + 2 x y ^{2} )}{( x ^{4} + y ^{2} ) ^{2}}

x^{4} + x^{3}, x^{4} + x^{2} + 6 x

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 0

\frac{d}{d x} (\frac{9 - x ^{2}}{8 x})

n = 1 \sum \infty \frac{( x ^{n} )}{7 n}

- 2 x

What is (\partial)/(\partial x)(x^2 y/(x^4+y^2)) ?
The answer to (\partial)/(\partial x)(x^2 y/(x^4+y^2)) is (y(-2x^5+2xy^2))/((x^4+y^2)^2)