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

The answer to (\partial)/(\partial x)(1/(3x^2+2y)) is -(6x)/((3x^2+2y)^2)

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

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

- \frac{6 x}{( 3 x ^{2} + 2 y ) ^{2}}

Solution steps

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

Treat

y

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{\partial}{\partial x} (3 x^{2} + 2 y) = 6 x

= - \frac{1}{( 3 x ^{2} + 2 y ) ^{2}} \cdot 6 x

Simplify

- \frac{1}{( 3 x ^{2} + 2 y ) ^{2}} \cdot 6 x : - \frac{6 x}{( 3 x ^{2} + 2 y ) ^{2}}

= - \frac{6 x}{( 3 x ^{2} + 2 y ) ^{2}}

\int 7 cos (π x) d x

\frac{\partial}{\partial x} (x^{3} y^{2} (6 - x - y))

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

\int sin^{2} (\frac{1}{3} x) d x

\frac{s}{( s ^{2} + 5 ^{2} ) ^{2}}

What is (\partial)/(\partial x)(1/(3x^2+2y)) ?
The answer to (\partial)/(\partial x)(1/(3x^2+2y)) is -(6x)/((3x^2+2y)^2)