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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 0 lim (\frac{x - x}{x})

\frac{d}{d x} ((x^{2} + 2)^{2})

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

\frac{5}{12} (3 + 5) x^{3}

\frac{\partial}{\partial v} (3 v cos (u))

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