What is (\partial)/(\partial y)(sqrt(x^3y-y^4)) ?

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

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(\partial)/(\partial y)(sqrt(x^3y-y^4))

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

\frac{x ^{3} - 4 y ^{3}}{2 x ^{3} y - y ^{4}}

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

\frac{1}{2 x ^{3} y - y ^{4}} \frac{\partial}{\partial y} (x^{3} y - y^{4})

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

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

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

Simplify

\frac{1}{2 x ^{3} y - y ^{4}} (x^{3} - 4 y^{3}) : \frac{x ^{3} - 4 y ^{3}}{2 x ^{3} y - y ^{4}}

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

\frac{d}{d x} (arcsin (9 x^{4}))

y^{^{''}} = 2 t^{2}

\frac{d}{d y} (\frac{6 x + y}{8 z - 5 y})

\frac{d}{d x} (- 4 ln (x))

\frac{\partial}{\partial x} (4 + cos (0.4 x - t))

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