What is (\partial)/(\partial x)(x/(x^6-y^3)) ?

The answer to (\partial)/(\partial x)(x/(x^6-y^3)) is (-5x^6-y^3)/((x^6-y^3)^2)

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(\partial)/(\partial x)(x/(x^6-y^3))

\frac{\partial}{\partial x} (\frac{x}{x ^{6} - y ^{3}})

\frac{- 5 x ^{6} - y ^{3}}{( x ^{6} - y ^{3} ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{x}{x ^{6} - y ^{3}})

Treat

y

as a constant

Apply the Quotient Rule:

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

= \frac{\frac{\partial}{\partial x} ( x ) ( x ^{6} - y ^{3} ) - \frac{\partial}{\partial x} ( x ^{6} - y ^{3} ) x}{( x ^{6} - y ^{3} ) ^{2}}

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

\frac{\partial}{\partial x} (x^{6} - y^{3}) = 6 x^{5}

= \frac{1 \cdot ( x ^{6} - y ^{3} ) - 6 x ^{5} x}{( x ^{6} - y ^{3} ) ^{2}}

1 \cdot (x^{6} - y^{3}) - 6 x^{5} x = - 5 x^{6} - y^{3}

= \frac{- 5 x ^{6} - y ^{3}}{( x ^{6} - y ^{3} ) ^{2}}

\frac{d y}{d x} = 4 x + y, y (0) = 9

\frac{d}{d x} (x^{16})

y^{^{''}} - 2 y^{^{'}} = sin (x)

\int x^{2} (2 x - 1)^{5} d x

\frac{d}{d x} (cos (x^{6}))

What is (\partial)/(\partial x)(x/(x^6-y^3)) ?
The answer to (\partial)/(\partial x)(x/(x^6-y^3)) is (-5x^6-y^3)/((x^6-y^3)^2)