What is (\partial)/(\partial x)(x/(x^4-y^6)) ?

The answer to (\partial)/(\partial x)(x/(x^4-y^6)) is (-3x^4-y^6)/((x^4-y^6)^2)

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(\partial)/(\partial x)(x/(x^4-y^6))

\frac{\partial}{\partial x} (\frac{x}{x ^{4} - y ^{6}})

\frac{- 3 x ^{4} - y ^{6}}{( x ^{4} - y ^{6} ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{x}{x ^{4} - y ^{6}})

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 ^{4} - y ^{6} ) - \frac{\partial}{\partial x} ( x ^{4} - y ^{6} ) x}{( x ^{4} - y ^{6} ) ^{2}}

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

\frac{\partial}{\partial x} (x^{4} - y^{6}) = 4 x^{3}

= \frac{1 \cdot ( x ^{4} - y ^{6} ) - 4 x ^{3} x}{( x ^{4} - y ^{6} ) ^{2}}

1 \cdot (x^{4} - y^{6}) - 4 x^{3} x = - 3 x^{4} - y^{6}

= \frac{- 3 x ^{4} - y ^{6}}{( x ^{4} - y ^{6} ) ^{2}}

\int cos^{2} (x) tan^{3} (x) d x

f (x) = \frac{e ^{x}}{x ^{2}}

x \to \frac{1}{4} lim (8 x (x - \frac{1}{5}))

x \to 0 lim (\frac{1}{1 - x})

y^{^{'}} = 0.5 (3 - y)

What is (\partial)/(\partial x)(x/(x^4-y^6)) ?
The answer to (\partial)/(\partial x)(x/(x^4-y^6)) is (-3x^4-y^6)/((x^4-y^6)^2)