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

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

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

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

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Popular >

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

Simplify

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

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

f (x) = 1 + sin (x)

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

\frac{\partial}{\partial x} (\frac{1}{2})

x y^{^{'}} - y = - 2 x^{2} + x - 2

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

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