What is the derivative of-(x^3/(y^3)) ?

The derivative of-(x^3/(y^3)) is -(3x^2(-x\frac{dy)/(dx)+y)}{y^4}

What is the first derivative of-(x^3/(y^3)) ?

The first derivative of-(x^3/(y^3)) is -(3x^2(-x\frac{dy)/(dx)+y)}{y^4}

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Cancel

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

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

\frac{d}{d x} (\frac{27}{x})

12 x

\int (5 x + 2)^{- 3} d x

\frac{d}{d x} (3 π x)

\int tan (4 x + 1) d x

What is the derivative of-(x^3/(y^3)) ?
The derivative of-(x^3/(y^3)) is -(3x^2(-x\frac{dy)/(dx)+y)}{y^4}
What is the first derivative of-(x^3/(y^3)) ?
The first derivative of-(x^3/(y^3)) is -(3x^2(-x\frac{dy)/(dx)+y)}{y^4}