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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y (x + 1) + y^{^{'}} = 0

\int \frac{cos ( θ )}{2 + sin ^{2} ( θ )} d θ

f (- 7) = \frac{1}{x + 3}

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