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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

x \to 0 lim (2 (1 + 3 x)^{\frac{2}{x}})

\frac{d}{d x} (\frac{3 x ^{2} + 2 x}{4 x - 2})

ma c l a u r in ln (3 + x)

\int \frac{1}{x ^{2} x ^{2} - 25} d x

d er i v a t i v e f (x) = 3 sin (x) + 3 cos (x)

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