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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{x - 3}{6 x - x ^{2}} d x

\frac{d}{d x} (ln (cos (x)))

\int \frac{tan ^{2} ( x )}{sec ^{5} ( x )} d x

\int \frac{w}{16 - w ^{2}} d w

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