What is the derivative of y=(2x)/(x+1) ?

The derivative of y=(2x)/(x+1) is 2/((x+1)^2)

What is the first derivative of y=(2x)/(x+1) ?

The first derivative of y=(2x)/(x+1) is 2/((x+1)^2)

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derivative of

y = \frac{2 x}{x + 1}

\frac{2}{( x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 x}{x + 1})

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x + 1) = 1

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

Simplify

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

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

7 (sin (t) y^{^{'}} + cos (t) y) = cos (t) sin^{3} (t)

\frac{d y}{d x} = \frac{- 2 x}{y}, y (0) = 1

\frac{d}{d x} (6 x + e^{- x})

(x^{2} y - x) d y + y d x = 0

\frac{d}{d x} (\frac{10}{x ^{7}})

What is the derivative of y=(2x)/(x+1) ?
The derivative of y=(2x)/(x+1) is 2/((x+1)^2)
What is the first derivative of y=(2x)/(x+1) ?
The first derivative of y=(2x)/(x+1) is 2/((x+1)^2)