What is the derivative of-(2x^3/y) ?

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

What is the first derivative of-(2x^3/y) ?

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

\int \frac{y}{( y + 2 ) ( 4 y - 1 )} d y

y = 4.5 x^{2} - 3 x + 2

\frac{\partial}{\partial x} (e^{\frac{x}{2}})

y^{^{'}} = sec^{2} (x) (1 - 3 y), y (\frac{π}{4}) = 0

f (x) = x^{2} + 16, at x = 3

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