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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{9}{1 + x ^{2}} d x

\frac{d}{d x} (\frac{3 x ^{2}}{ln ( x )})

\frac{d y}{d x} = (x + y)^{2} - 1, y (3) = 4

\int_{0}^{1} 2 (1 + x)^{8} d x

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

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