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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{e ^{2 x}}{( 1 + e ^{2 x} ) ^{3}} d x

\int (x^{2} + y^{2})^{\frac{3}{2}} d x

n = 0 \sum \infty \frac{4 ^{n}}{4 ^{n}}

\frac{d}{d x} (lo g_{e} (x - 1))

\frac{d}{d x} (\frac{1}{4 + x})

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