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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

\frac{\partial}{\partial x} (sin (- 2 x) cos (4 y))

\frac{d}{d x} (x^{2} + a \cdot x + b)

\frac{d y}{d x} = x^{2} (1 + y)

\frac{d}{d x} (ln (\frac{x + 3}{x - 3}))

\int \frac{1}{x 8 - ln ^{2} ( x ) + 2 ln ( x )} d x

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