What is the derivative of y= x/(2x+\frac{-4){x}} ?

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

What is the first derivative of y= x/(2x+\frac{-4){x}} ?

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

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derivative of

y = \frac{x}{2 x + \frac{- 4}{x}}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

= \frac{1 \cdot ( 2 x + \frac{- 4}{x} ) - ( 2 + \frac{4}{x ^{2}} ) x}{( 2 x + \frac{- 4}{x} ) ^{2}}

Simplify

\frac{1 \cdot ( 2 x + \frac{- 4}{x} ) - ( 2 + \frac{4}{x ^{2}} ) x}{( 2 x + \frac{- 4}{x} ) ^{2}} : - \frac{2 x}{( x ^{2} - 2 ) ^{2}}

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

9 y y^{^{'}} + 4 x = 0

e^{c o s (x)}

\frac{d}{d x} (csc^{9} (x))

\frac{4 x}{x + 2} (2.2)

\frac{1}{10 x}, at x = 9

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