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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

\frac{d y}{d x} = - \frac{( x + 6 )}{7 y ^{2}}

\frac{d θ}{d t} = 40 - \frac{2}{5} (θ - 15)

y^{^{''}} + 4 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = 1

f (x) = (9 x^{2} - 15 x) e^{x}

\int sin (x) cos (x) ln (sin (x)) d x

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