What is (\partial)/(\partial x)(x/(x^4-y^5)) ?

The answer to (\partial)/(\partial x)(x/(x^4-y^5)) is (-3x^4-y^5)/((x^4-y^5)^2)

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(\partial)/(\partial x)(x/(x^4-y^5))

\frac{\partial}{\partial x} (\frac{x}{x ^{4} - y ^{5}})

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

Solution steps

\frac{\partial}{\partial x} (\frac{x}{x ^{4} - y ^{5}})

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 ^{5} ) - \frac{\partial}{\partial x} ( x ^{4} - y ^{5} ) x}{( x ^{4} - y ^{5} ) ^{2}}

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

\frac{\partial}{\partial x} (x^{4} - y^{5}) = 4 x^{3}

= \frac{1 \cdot ( x ^{4} - y ^{5} ) - 4 x ^{3} x}{( x ^{4} - y ^{5} ) ^{2}}

1 \cdot (x^{4} - y^{5}) - 4 x^{3} x = - 3 x^{4} - y^{5}

= \frac{- 3 x ^{4} - y ^{5}}{( x ^{4} - y ^{5} ) ^{2}}

f^{8}

\int x + 9 d x

\int_{- 2}^{1} (x - 4 ∣ x ∣) d x

y^{^{''}} - 4 y = e^{- 2 x} - 2 x

What is (\partial)/(\partial x)(x/(x^4-y^5)) ?
The answer to (\partial)/(\partial x)(x/(x^4-y^5)) is (-3x^4-y^5)/((x^4-y^5)^2)