What is (\partial)/(\partial x)(1/(x^2+3y)) ?

The answer to (\partial)/(\partial x)(1/(x^2+3y)) is -(2x)/((x^2+3y)^2)

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(\partial)/(\partial x)(1/(x^2+3y))

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + 3 y})

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

Solution steps

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + 3 y})

Treat

y

as a constant

Apply exponent rule:

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

= \frac{\partial}{\partial x} ((x^{2} + 3 y)^{- 1})

Apply the chain rule:

- \frac{1}{( x ^{2} + 3 y ) ^{2}} \frac{\partial}{\partial x} (x^{2} + 3 y)

= - \frac{1}{( x ^{2} + 3 y ) ^{2}} \frac{\partial}{\partial x} (x^{2} + 3 y)

\frac{\partial}{\partial x} (x^{2} + 3 y) = 2 x

= - \frac{1}{( x ^{2} + 3 y ) ^{2}} \cdot 2 x

Simplify

- \frac{1}{( x ^{2} + 3 y ) ^{2}} \cdot 2 x : - \frac{2 x}{( x ^{2} + 3 y ) ^{2}}

= - \frac{2 x}{( x ^{2} + 3 y ) ^{2}}

\int \frac{x ^{2} + x}{x ^{3} - x ^{2} + x - 1} d x

y^{^{'}} + \frac{1 - x}{2 x} y = 0

f (x) = 2 x + \frac{1936}{x}

f (x) = 110 x - 16 x^{2}, at x = 4

(x^{2} - x y) d x + x d y = 0

What is (\partial)/(\partial x)(1/(x^2+3y)) ?
The answer to (\partial)/(\partial x)(1/(x^2+3y)) is -(2x)/((x^2+3y)^2)