What is (\partial)/(\partial x)((x^3+5)/(y^3-1)) ?

The answer to (\partial)/(\partial x)((x^3+5)/(y^3-1)) is (3x^2)/(y^3-1)

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(\partial)/(\partial x)((x^3+5)/(y^3-1))

\frac{\partial}{\partial x} (\frac{x ^{3} + 5}{y ^{3} - 1})

\frac{3 x ^{2}}{y ^{3} - 1}

Solution steps

\frac{\partial}{\partial x} (\frac{x ^{3} + 5}{y ^{3} - 1})

Treat

y

as a constant

Take the constant out:

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

= \frac{1}{y ^{3} - 1} \frac{\partial}{\partial x} (x^{3} + 5)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{1}{y ^{3} - 1} (\frac{\partial}{\partial x} (x^{3}) + \frac{\partial}{\partial x} (5))

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

\frac{\partial}{\partial x} (5) = 0

= \frac{1}{y ^{3} - 1} (3 x^{2} + 0)

Simplify

\frac{1}{y ^{3} - 1} (3 x^{2} + 0) : \frac{3 x ^{2}}{y ^{3} - 1}

= \frac{3 x ^{2}}{y ^{3} - 1}

\frac{d}{d x} (arccos (e^{3 x}))

\frac{1}{x ^{3}}

\int sec (x) tan (x) - 4 x d x

\int (x^{4} + 5)^{3} x^{3} d x

\frac{d}{d x} (cos^{2} (x) sin^{2} (x))

What is (\partial)/(\partial x)((x^3+5)/(y^3-1)) ?
The answer to (\partial)/(\partial x)((x^3+5)/(y^3-1)) is (3x^2)/(y^3-1)