What is (\partial)/(\partial x)(2/(2x-y)) ?

The answer to (\partial)/(\partial x)(2/(2x-y)) is -4/((2x-y)^2)

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(\partial)/(\partial x)(2/(2x-y))

\frac{\partial}{\partial x} (\frac{2}{2 x - y})

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

Solution steps

\frac{\partial}{\partial x} (\frac{2}{2 x - y})

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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5^{t}

f (x) = x + 3, at x = 1

\frac{d}{d x} (\frac{ln ( x ^{5} )}{ln ( x ^{4} )})

\int_{- 2}^{2} (8 - 2 y^{2}) d y

What is (\partial)/(\partial x)(2/(2x-y)) ?
The answer to (\partial)/(\partial x)(2/(2x-y)) is -4/((2x-y)^2)