What is (\partial)/(\partial x)((7xy)/(x^2+y^4)) ?

The answer to (\partial)/(\partial x)((7xy)/(x^2+y^4)) is (7y(-x^2+y^4))/((x^2+y^4)^2)

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(\partial)/(\partial x)((7xy)/(x^2+y^4))

\frac{\partial}{\partial x} (\frac{7 x y}{x ^{2} + y ^{4}})

\frac{7 y ( - x ^{2} + y ^{4} )}{( x ^{2} + y ^{4} ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{7 x y}{x ^{2} + y ^{4}})

Treat

y

as a constant

Take the constant out:

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

= 7 y \frac{\partial}{\partial x} (\frac{x}{x ^{2} + y ^{4}})

Apply the Quotient Rule:

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

= 7 y \frac{\frac{\partial}{\partial x} ( x ) ( x ^{2} + y ^{4} ) - \frac{\partial}{\partial x} ( x ^{2} + y ^{4} ) x}{( x ^{2} + y ^{4} ) ^{2}}

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

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

= 7 y \frac{1 \cdot ( x ^{2} + y ^{4} ) - 2 xx}{( x ^{2} + y ^{4} ) ^{2}}

Simplify

7 y \frac{1 \cdot ( x ^{2} + y ^{4} ) - 2 xx}{( x ^{2} + y ^{4} ) ^{2}} : \frac{7 y ( - x ^{2} + y ^{4} )}{( x ^{2} + y ^{4} ) ^{2}}

= \frac{7 y ( - x ^{2} + y ^{4} )}{( x ^{2} + y ^{4} ) ^{2}}

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

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\frac{\partial}{\partial z} (- 3 z)

What is (\partial)/(\partial x)((7xy)/(x^2+y^4)) ?
The answer to (\partial)/(\partial x)((7xy)/(x^2+y^4)) is (7y(-x^2+y^4))/((x^2+y^4)^2)