What is the derivative of (2x/(x^2+y^4)) ?

The derivative of (2x/(x^2+y^4)) is (2(-x^2-4xy^3\frac{dy)/(dx)+y^4)}{(x^2+y^4)^2}

What is the first derivative of (2x/(x^2+y^4)) ?

The first derivative of (2x/(x^2+y^4)) is (2(-x^2-4xy^3\frac{dy)/(dx)+y^4)}{(x^2+y^4)^2}

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

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

\frac{\partial}{\partial x} (x^{2} + 8 y^{2} + 9 z^{2})

y^{^{''}} - 2 y^{^{'}} - 3 y = - 7 t e^{- t}

ln (cos (x) + sin (x))

\int tan^{4} (x co) s^{5} x

What is the derivative of (2x/(x^2+y^4)) ?
The derivative of (2x/(x^2+y^4)) is (2(-x^2-4xy^3\frac{dy)/(dx)+y^4)}{(x^2+y^4)^2}
What is the first derivative of (2x/(x^2+y^4)) ?
The first derivative of (2x/(x^2+y^4)) is (2(-x^2-4xy^3\frac{dy)/(dx)+y^4)}{(x^2+y^4)^2}