What is (\partial)/(\partial x)(1/(x^2+4y^2)) ?

The answer to (\partial)/(\partial x)(1/(x^2+4y^2)) is -(2x)/((x^2+4y^2)^2)

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(\partial)/(\partial x)(1/(x^2+4y^2))

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + 4 y ^{2}})

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

Solution steps

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + 4 y ^{2}})

Treat

y

as a constant

Apply exponent rule:

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

= \frac{\partial}{\partial x} ((x^{2} + 4 y^{2})^{- 1})

Apply the chain rule:

- \frac{1}{( x ^{2} + 4 y ^{2} ) ^{2}} \frac{\partial}{\partial x} (x^{2} + 4 y^{2})

= - \frac{1}{( x ^{2} + 4 y ^{2} ) ^{2}} \frac{\partial}{\partial x} (x^{2} + 4 y^{2})

\frac{\partial}{\partial x} (x^{2} + 4 y^{2}) = 2 x

= - \frac{1}{( x ^{2} + 4 y ^{2} ) ^{2}} \cdot 2 x

Simplify

- \frac{1}{( x ^{2} + 4 y ^{2} ) ^{2}} \cdot 2 x : - \frac{2 x}{( x ^{2} + 4 y ^{2} ) ^{2}}

= - \frac{2 x}{( x ^{2} + 4 y ^{2} ) ^{2}}

\frac{d y}{d x} = 4 x + y

y^{^{'}} = y (7 - y) - \frac{49}{4}

\frac{d}{d x} (3 x^{3} - 5 x)

\frac{2}{x ^{2}} + \frac{3}{x ^{3}}

x \to 2 - lim (\frac{x ^{2}}{( x - 2 )})

What is (\partial)/(\partial x)(1/(x^2+4y^2)) ?
The answer to (\partial)/(\partial x)(1/(x^2+4y^2)) is -(2x)/((x^2+4y^2)^2)