What is (\partial)/(\partial x)(2x(2+y)^{-1}) ?

The answer to (\partial)/(\partial x)(2x(2+y)^{-1}) is 2/(2+y)

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(\partial)/(\partial x)(2x(2+y)^{-1})

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

\frac{2}{2 + y}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the common derivative:

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

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

Simplify

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

= \frac{2}{2 + y}

n = 1 \sum \infty 9 - \frac{6}{n ^{2}}

x \to 1 lim (- x^{2} + 1)

\int 7 e^{8 x} d x

n = 0 \sum \infty ln^{n} (3)

t y^{^{'}} + (t + 1) y = t, y (ln (5)) = 1

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