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

The derivative of (4y^2-x^4/(4xy)) is (x^5\frac{dy)/(dx)-3x^4y+4xy^2(dy)/(dx)-4y^3}{4x^2y^2}

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

The first derivative of (4y^2-x^4/(4xy)) is (x^5\frac{dy)/(dx)-3x^4y+4xy^2(dy)/(dx)-4y^3}{4x^2y^2}

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Popular >

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

\frac{x ^{5} \frac{d y}{d x} - 3 x ^{4} y + 4 x y ^{2} \frac{d y}{d x} - 4 y ^{3}}{4 x ^{2} y ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (4 y^{2} - x^{4}) = 8 y \frac{d y}{d x} - 4 x^{3}

\frac{d}{d x} (x y) = y + x \frac{d y}{d x}

= \frac{1}{4} \cdot \frac{( 8 y \frac{d y}{d x} - 4 x ^{3} ) x y - ( y + x \frac{d y}{d x} ) ( 4 y ^{2} - x ^{4} )}{( x y ) ^{2}}

Simplify

\frac{1}{4} \cdot \frac{( 8 y \frac{d y}{d x} - 4 x ^{3} ) x y - ( y + x \frac{d y}{d x} ) ( 4 y ^{2} - x ^{4} )}{( x y ) ^{2}} : \frac{x ^{5} \frac{d y}{d x} - 3 x ^{4} y + 4 x y ^{2} \frac{d y}{d x} - 4 y ^{3}}{4 x ^{2} y ^{2}}

= \frac{x ^{5} \frac{d y}{d x} - 3 x ^{4} y + 4 x y ^{2} \frac{d y}{d x} - 4 y ^{3}}{4 x ^{2} y ^{2}}

\int \frac{2 x}{x ^{2} - 49} d x

ln (3 - x), 2

\int \frac{sec ^{2} ( x )}{x} d x

\int cos (nπ x) d x

\frac{d}{d x} (arccos (\frac{1 - x ^{2}}{1 + x ^{2}}))

What is the derivative of (4y^2-x^4/(4xy)) ?
The derivative of (4y^2-x^4/(4xy)) is (x^5\frac{dy)/(dx)-3x^4y+4xy^2(dy)/(dx)-4y^3}{4x^2y^2}
What is the first derivative of (4y^2-x^4/(4xy)) ?
The first derivative of (4y^2-x^4/(4xy)) is (x^5\frac{dy)/(dx)-3x^4y+4xy^2(dy)/(dx)-4y^3}{4x^2y^2}