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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

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

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

\int \frac{1}{sec ^{4} ( \frac{1}{4} β )} d β

\int 6 \cdot x + 4 d x

\frac{\partial}{\partial x} (sin (x + 2 y + 3 z))

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

x \to 4 lim (\frac{7}{( x - 4 )})

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