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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

f (x) = - 3 x^{2} - 2 x, at x = 2

n = 0 \sum \infty (\frac{4}{10})^{n}

\int_{2}^{3} 21 t^{2} d t

n = 1 \sum \infty \frac{n ^{4} x ^{3 n}}{5 ^{3 n}}

\frac{d}{d x} (4 x + 12 y, 12 x - 3 y)

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