What is (\partial)/(\partial x)((2x)/(x+1)) ?

The answer to (\partial)/(\partial x)((2x)/(x+1)) is 2/((x+1)^2)

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(\partial)/(\partial x)((2x)/(x+1))

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

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

Solution steps

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

Take the constant out:

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

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

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

\frac{\partial}{\partial 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}}

Y (θ) = cot^{2} (cos (θ))

f (t) = 6 t - 5 t^{2}

\frac{\partial}{\partial x} (e^{- 4 x - x^{2} + 2 y - y^{2}})

\frac{1}{x + 2}

x \to + 0 + lim (\frac{x ^{2}}{6} - \frac{5}{x})

What is (\partial)/(\partial x)((2x)/(x+1)) ?
The answer to (\partial)/(\partial x)((2x)/(x+1)) is 2/((x+1)^2)