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

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

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

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

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

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

Solution steps

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

Simplify

\frac{- 2 x}{2 y} : - \frac{x}{y}

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

Simplify

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

\frac{d}{d x} (\frac{h x}{2} + \frac{k d}{x})

\frac{\partial}{\partial x} (x \cdot e^{\frac{y}{x}})

f^{^{'}} (x) = \frac{4}{1 - 2 x ^{2}}

sin (t^{2}) ln (t^{3})

\int (e^{\frac{x}{a}} - e^{- \frac{x}{a}})^{2} d x

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