What is (\partial)/(\partial x)((x^3-y^2)^{(-1)}) ?

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

3 y^{^{''}} - 4 y^{^{'}} - 15 y = 0, y (0) = - 7, y^{^{'}} (0) = - 2

\frac{d}{d x} ((9 x^{2} - 18 x + 18) e^{x})

\frac{d}{d x} ((x - 1) (x + 1) (ln (900) - ln (x)))

\frac{1}{s ( s ^{2} + 9 )}

f (x) = 2 cos (x) + cos^{2} (x)

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